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0:01
Welcome everyone. We're going to explore some really fundamental probability distributions. First up, problem one, website conversion. Imagine a potential customer visits your e-commerce website. There's a 3% chance they will make a purchase. So, part A asks, which distribution models whether this specific customer completes a purchase? Let's think about the clues here. We're looking at a single customer and there are really only two outcomes, right?
0:30
They either buy or they don't. That's right. It's the Berni distribution. It's the simplest one, perfect for a single trial with a binary outcome. We can say X= 1 for a purchase, which happens with a probability P of 0.03, and X= 0 for no purchase with a probability of 1 minus P, which is 0.97. Simple as that. Okay, so for part B, what is the expected value or the mean for this situation?
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And what's the probability they don't make a purchase? Here's the great thing about Berni. The formula is super easy. The expected value for a Bernoli distribution is simply P. So in our case, the expected value E of X is 0.03. And the probability of no purchase, well, that's just 1 minus P. So we do 1 - 0.0 03 which gives us 0.97 or 97%. Fantastic. Let's move on to problem two. Customer service calls. A call center receives calls at an average rate of 15 calls per hour. The
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question for part A is which distribution models the number of calls received in a 15minute period. The key hints here are that we're counting random events over a fixed time period. Does a specific distribution come to mind? You got it. It's the plus distribution. It's perfect for this. Now, here's the first step, and it's an important one. We have to scale the rate, which we call lambda
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for our 15-minute period. The original rate is 15 calls per hour. So, we take 15 and multiply it by 15 minutes over 60 minutes per hour. That gives us a lambda of 3.75 calls per 15 minutes. As you can see on the slide, the probability of getting k calls is given by that classic pson formula. Now for part B, given our rate, lambda is 3.75 per 15 minutes. What is the probability that exactly
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three calls are received in a 15-minute period? Okay, let's walk through this formula together. We need to calculate P of X= 3 using our lambda of 3.75. So we plug the numbers into the formula you see on the screen. That's 3.75 to the^ of 3 * e the^ of -3.75 all divided by 3 factorial. Let's break that down. 3.75 cubed is about 52.7. E to the -3.75 is about 0.0235. 0235 and 3 factorial is just three six. When you crunch all those numbers, you get approximately 0.2066.
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So there's about a 21% chance of getting exactly three calls. Not bad. Problem three, the die roll. A classic. A standard six-sided die is rolled once. Part A asks, which distribution models the number that appears on top. Think about it. Are all the outcomes equally likely? The name starts with a U. Of course, it's the discrete uniform distribution. This one's straightforward. Each of the six outcomes 1 2 3 4 5 or six has an equal probability.
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So P of X= K is simply 1 / 6 for any of those numbers. Okay. Part B. If the probability for any single number is 16th, what is the probability of rolling a two, a three, or a four? Well, since these events are mutually exclusive, you can't roll a two and a three at the same time. We can just add their probabilities together. So we take the probability of rolling a two which is 16th plus the probability of a three another 16th plus the probability of a 4 which is you guessed
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it 16th. That gives us 3 / 6 which simplifies to 1/2 or 50%. Easy peasy. Now for a more complex one. Problem four. Income distribution. Household income data is often right skewed. This means most people earn lower to middle incomes and there's a long tale of very high earners. So part A is which distribution often models this kind of data. Here's a hint. When you take the log of something that's right skewed, it often becomes normal. This is the lognormal distribution.
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It's perfect for modeling positive values that tend to grow multiplicatively like income or stock prices. The key idea is this. If the income X is lognormally distributed, then the natural log of X follows a normal distribution. You can see the notation there. Ln of X is distributed normally with mean mu and variance sigma squared. Okay, this next part has a few steps. So let's walk through it carefully together. For part B, let's assume the natural log of income is
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normally distributed with a mean, we'll call it mu sublog of 11.2 and a standard deviation sigma sublog of 0.7. The question is, what percentage of households earn more than $150,000? Okay, we are going to walk through this entire process together step by step. Step one, we can't work with the dollar amount directly. We have to transform the income threshold to the log scale. So, we take the natural log of 150,000,
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which is approximately 11.918. Step two, now we calculate the zcore. Remember that formula? It's our value minus the mean divided by the standard deviation. So, we take 11.918, subtract the mean of 11.2, 2 and divide the result by the standard deviation of 0.7. This gives us a zcore of about 1.026. Step three, find the probability. We want the probability that Z is greater than 1.026. A standard Z table gives us the area to the left. So, we do 1 minus
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the probability that Z is less than or equal to 1.02. 26. Looking that up, we get 1us 0.8474, which equals 0.1526. The final result, 15.26%. So about 15% of households earn more than $150,000 in this model. Great work getting through that one. Problem five, waiting for a call.
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This ties back to problem two. A call center receives calls at an average rate of 15 calls per hour, which we know is a plus process. So for part A, which distribution models the time until the next call? The hints are if arrivals are pasaw, what describes the waiting time? And this distribution has the famous memoryless property. It's the exponential distribution. It's the flip side of the posson coin. Pson counts the number of events in a time period. Exponential measures
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the waiting time between those events. You can see the formula for the probability that the wait time x is less than or equal to sometime t on the screen. Okay, let's finish this. Part B. A call center gets 15 calls per hour. What is the probability that the time until the next call is less than 5 minutes? Let's break it down. Step one, we need our units to match. The rate lambda is in calls per hour, but our time t is in minutes. So, let's convert the rate. 15 calls
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per 60 minutes gives us a lambda of 0.25 calls per minute. Step two, use the formula which is called the cumulative distribution function or CDF with t= 5 minutes. The probability that x is less than 5 is 1 - e ^ of lambda * t. So that's 1 - e ^ of0.25 * 5. This becomes 1 - e ^ of -1.25. And when you calculate that, you get 1us 0.2865, which is 0.7135. The final answer is 71.35%. There's a pretty high chance the next call will come within 5 minutes. Problem six,
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coin flips. Imagine this. You flip a fair coin 10 times and you want to count the number of heads. So for part A, which distribution models this scenario? The key hints are that we're counting successes in a fixed number of trials. [Music] If you guess the binomial distribution, you are absolutely correct. We have n= 10 independent trials, right? And the success probability P for a fair coin is 0.5. The formula for getting exactly k heads is right there on the
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screen. It's c of 10 comma k * 0.5 ^ of k * 0.5 ^ of 10 - k which simplifies nicely to c of 10 comma k * 0.5 ^ of 10. Okay. So for part B, what is the probability of getting exactly seven heads? Let's use our formula. We're using the binomial formula with n= 10, k= 7, and p = 0.5. So the probability that our random variable X= 7 is C of 10 7 * 0.5 to the power of 7 * 0.5 to the power of 3. Now 10 choose 7 is 120. We multiply that by 0.78125
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and then by 0.125. And the result is 0.1172 or 11.72%. Not bad at all. Let's move on to problem seven, first sale. A salesperson has a 15% chance of closing a deal on any given call. Part A asks, which distribution models the number of calls until their first successful sale? The big clue here is that we are waiting for the first success first. And here's a little hint. Think geometry. This is the geometric distribution. It's perfect for modeling the
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number of trials K needed for that very first success. The probability of the first success happening on trial K is given by the formula you see here. P of X equ= K is 1 - P to the power of K -1 * P. For our salesperson, that's 0.85 to the power of K minus1 * 0.15. So for part B, given that a salesperson has a 15% chance of closing a deal, what is the probability that their first sale occurs on their third call? Let's think about what has to happen. This means they
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have two failures followed by one success. So we plug K = 3 into our formula. P of X = 3 is 1 - P to the power of 3 - 1 * P. That's 0.85^ 85^ 2AR * 0.15 which is 0.7225 * 0.15 giving us 10.84%. So there's about an 11% chance their first win is on call number three. On to problem eight. Dice rolls with categories. You roll a six-sided die 12 times and you count how many ones, how many twos,
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and how many other numbers you get. So, part A, which distribution models this scenario? The hints are we have more than two outcomes but still a fixed number of trials? This is a job for the multinnomial distribution. It's basically the big brother of the binomial distribution handling more than two categories. Here our probabilities are P1 the probability of rolling a one is 16th. P2
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rolling a two is also 16th and P3 rolling other is 46ths. The formula looks a little intense I know but it's just counting all the combinations and multiplying by their probabilities. Part B. With those probabilities and a total of n equals 12 rolls, what's the probability of getting five 1's, 3 2's, and four other outcomes? Select. Okay, we need to find the probability that x1 is 5, x2 is 3, and x3 is 4. Let's plug everything into that big formula. We have 12 factorial / 5
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factorial * 3 factorial * 4 factorial. All of that is then multiplied by 16 to the power of 5 * 1 16 to the^ of 3 * 4 6 to the power of 4. Okay, that first part the big fraction simplifies to 27,720. We multiply that by all those tiny probabilities and we get approximately 0.00326. Wow. The result is 0.33%. That's a very specific and rare outcome. Problem nine, multiple sales. A cold caller has a 10% chance of making a sale on any call. Part A. Which distribution models the
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number of calls needed to achieve three sales? Notice the difference here. It's not the first success, but a fixed number of successes. We're waiting for success number three. This is the negative binomial distribution. It's designed specifically to model the number of trials K you need to get a fixed number of successes, R. For our case with R= 3 sales and P= 0.1, you can see the probability formula right on the screen. It's C of K -1, R -1 * P R * 1 - P to the K minus
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R. So let's ask a slightly different question for part B. To achieve R equ= 3 sales with that success probability of 10%, how many calls would they expect to make? This is actually much simpler than it sounds. The expected value or the mean of a negative binomial distribution is just mu= r / p. So for us that's 3 / 0.10 which equals 30. They should expect to make 30 calls to get those three sales. It makes perfect sense, doesn't it? Last one. Problem 10. Heights of adults.
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We know that most adult women are of average height with fewer being very short or very tall. The data forms a smooth unimodal curve. Part A. Which distribution would be appropriate? The biggest hint is it's the most famous bell curve. Of course, it's the normal distribution. It models so many natural phenomena where values cluster symmetrically around a mean, which we call mu.
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Probabilities are found using the standard normal cumulative distribution function which is that capital PH symbol after converting our value to a Zcore. You can see the formula right there. P of X is less than or equal to X= 5 ofx - mu over sigma. Okay, let's use it. Part B. Male heights are normal with a mean mu of 175 cm and a standard deviation sigma of 6 cm. What is the probability a male is between 170 cm and 185 cm tall? Let's walk through
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this together step by step. Step one, find the zcore for 170 cm. The formula is value minus mean divided by standard deviation. So 170 minus 175 / 6. That gives us approximately0.83. Step two, find the zcore for 185 cm. That's 185us 175 / 6, which is about 1.67. Step three. Now we find the probability that Z is between0.83 and 1.67. The way we do this is we find the probability that Z is less than 1.67 and subtract the probability that Z is less than0.83.
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Looking these up in a Z table, we get 0.9525 minus 0.2033. 2033 and the result is 74.92%. And that's it. That's how you solve it. Problem 11, birthday selection. The problem is simple. A person's birthday can fall on any of the 12 months of the year. So for part A, which distribution models the month in which someone is born? The key question to ask yourself is on the slide. Are all 12 months equally likely? If we assume that births are spread out evenly, then yes. Each
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month has the exact same chance. This means we're dealing with a discrete uniform distribution. For a random variable X, which represents the month number, the probability of being born in any specific month K is simply 1 / 12. Easy as that. Okay, let's build on that. For part B, what is the probability that someone is born in the first quarter? That would be January, February, or March. Well, since each month has a 1 in12 probability and the first quarter
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includes three months, we just add them up. The probability of the first quarter is the probability of January plus the probability of February plus the probability of March. That's 112th + 112th + 112th which equals 3 12ths. And that simplifies down to a result of 1/4 or 25%. Okay, on to problem 12. A marketing campaign. An email campaign is sent to 20 subscribers. The open rate is consistently 25%. So part A, which distribution models the number of subscribers who
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open the newsletter? Take a look at the hint on the slide. We are counting successes in a fixed number of trials. That's right. It's the binomial distribution. It's perfect for this scenario. We have n equals 20 trials which are our subscribers and each trial has a success probability p of 0.25 which is the chance they open the email. You can see the formula on the screen for the probability of exactly k people opening. It's c of 20 k * 0.25 25
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to the K * 0.75 to the 20 minus K. So for part B, let's use that formula. What is the probability that exactly five subscribers will open the newsletter? Let's walk through this. We have our binomial distribution with n= 20 and P= 0.25. We need to calculate P of X= 5.
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Looking at the formula, that means we calculate C of 20, 5, which comes out to 15,54. Then we multiply that by 0.25 to the power of 5 and then by 0.75 to the power of 15. It's a bit of a calculation, but when the dust settles, the result is approximately 0.2023. That's a 20.23% chance. Next up, problem 13. Machine lifetime. An electronic components failure probability is constant no matter its age. Part A. Which distribution models
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the component's lifetime? The critical hint is right there. The component has no memory of how long it's been running. That key feature points directly to the exponential distribution. It's used to model the time until an event occurs in a process where the event rate is constant. The probability that the component lasts longer than some time t is given by the formula e to the power of negative lambda t where lambda is the failure rate. Now for part b,
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the component follows an exponential distribution with an average lifetime of 10,000 hours. What's the probability it will last longer than 12,000 hours? Let's break this down into two simple steps. Step one, we need to find the failure rate lambda. For an exponential distribution, the mean is just 1 / lambda. So lambda equals 1 / 10,000 which is 0.001 failures per hour. Okay. Now for step two, we calculate the probability that X is greater
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than 12,000. We just plug our numbers into the formula. It's e to the power of0.001 * 12,000. That simplifies to e to the power of -1.2, which is approximately 03012. So the result is a 30.12% chance. Problem 14, stock prices. A stock price can't be negative and its future value is often right skewed because of compounding gains. Part A. What distribution is commonly used to model stock prices? The hint is a big one. If the
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logarithm of your data looks normal, then you're dealing with a lognormal distribution. It's a really important one in finance. A random variable X is lognormally distributed if its natural logarithm ln of X is normally distributed. So if X is our stock price, then the natural log of X follows a normal distribution. Okay, let's apply this part B. The log of a future stock price is normal with a mean of 3.8 and a standard deviation of 0.2. What's
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the probability the actual stock price will be less than $40? We are going to walk through this entire process together step by step. First step one, we have to transform our price threshold of $40 to the log scale. The natural log of 40 is approximately 3.689. Step two, now we calculate the zcore for that log value. The formula is value minus mean divided by standard deviation. So 3.689us 3.8 all divided by 0.2 two. This gives us a zcore of about0.555.
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Finally, step three, we just need to find the probability that Z is less than0.555 from a standard Z table and the result is 28.9%.
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Problem 15, the free throw. A basketball player attempts one free throw shot. Part A. Which distribution models whether the shot goes in? The hints are key. Just one trial, two outcomes. This is the simplest one of all. It's the Bernui distribution. It's for a single experiment with only two possible outcomes, success or failure. We say x= 1 if they make the shot with probability p and x= 0 if they miss with probability 1 minus p. Part B. This player makes
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75% of their free throws. So P equals 0.75. The question is, what's the probability they make this shot and what's the expected value? This is almost a trick question because it's so straightforward. The probability of success is simply P itself. So, the probability they make the shot is 0.75. And here's the cool part. The expected value of a Bernoli trial is also P. So the expected value is
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also 0.75. It's that simple. Great job on this set. Problem 16. Finding a defect. An inspector checks items from an assembly line where 2% of items are defective. First question, which distribution models the number of items checked until finding the first defective item? The key here is that we're counting trials until the first success. That my friends is the geometric distribution. This models the number of trials K needed for the very first success. In our case,
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a success is finding a defect. So P is 0.02. The probability of that first defect showing up on the kith item is P of X = K = 1 - P to the power of K -1 * P. which for us is 0.98 to the k minus one * 0.02. So part B, the probability of a defect is 0.02. How many items on average would the inspector expect to check before finding that first defective one? Here's the great thing about the geometric distribution. The expected value or the mean is just 1 / p. It's super intuitive.
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So, mu = 1 / 0.02, which is exactly 50. The inspector expects to check 50 items. Let's move on to problem 17. Website page views. A news article gets 50 page views per minute on average. Part A. Which distribution models the number of page views in a six-second window?
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H. We're counting events in a fixed interval. That's the telltale sign of a posson distribution. But first, we have a little prep work to do. We need to scale the rate lambda for our 6-second window. A 6-second window is 1/10enth of a minute. So, we take 50 views per minute and multiply by 1 minute per 10 windows, giving us a lambda of 5 views per 6 seconds. Now we can use the formula. The probability of K views is lambda to the K * E to the negative
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lambda all divided by K factorial. Okay. Part B. The rate lambda is five page views per 6cond window. What's the probability that it receives no page views in a specific six-second window?
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We need to calculate P of X= 0 using our lambda of 5. So it's 5 to the power of 0 power of 0 * e to the - 5 all divided by 0 factorial. And remember anything to the power of 0 is 1 and 0 factorial is also 1. So it just simplifies down to e to the -5 which is approximately 0.67. That's a tiny probability just 0.67%. Problem 18. The random number generator. A computer's random number generator produces real numbers between zero and one. So for part A, which distribution models these
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generated numbers? Let me ask you this. Is any number more likely than another? Nope. Every number between 0 and one has an equal chance of being selected. This is a uniform distribution. The continuous version. The probability that our number x falls between some value a and b is just b minus a for any valid interval between 0 and 1. It's beautifully simple. Which brings us to part b. Numbers are generated uniformly
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between 0 and 1. What's the probability a generated number falls between 0.3 and 0.7? Just like the formula said, the probability is simply the width of the interval. So P of X being between 0.3 and 0.7 is just 0.7 minus 0.3. The result is 0.4 or 40%. Problem 19. Basketball shots. A player makes 70% of her free throws. We're interested in the number of attempts to score her fifth basket.
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Part A. Which distribution models this? Notice we're not waiting for the first success anymore. We're waiting for the arth success. This is a job for the negative binomial distribution. It's the perfect tool for modeling the number of trials K to get a fixed number of successes R. For R equals five baskets and a probability of P= 0.7. The formula looks a little intense, but it's not so bad. P of X= K is K minus one choose R minus one * P R * 1 - P to the K minus
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R. Let's use it. Part B. We need the fifth basket. So R equ= 5 with a success probability of 0.7. What's the probability that she scores her fifth basket on her seventh attempt? We are going to walk through this calculation step by step. We need to calculate P of X = 7. So, we plug in our numbers. That's 7 - 1 choose 5 - 1 * 0.7 power * 0.3 to the power of 7 - 1 5. Let's simplify that. It becomes 6 choose 4 * 0.7 5th * 0.3 squared. 6 choose 4 is 15. 0.7 5th is
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about 0.168 and 0.3^2ar is 0.09. Multiply those all together and you get approximately 0.2269. The result 22.69%. Problem 20. Customer preferences. A restaurant observes customer orders 60% coffee, 30% tea, and 10% juice. We select 10 customers. Part A. Which distribution models the number ordering each beverage? It feels like binomial but with more than two categories. And that's exactly what it is. It's the multinnomial distribution. It generalizes the binomial for n trials
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with c possible outcomes. In our case, n is 10 customers. And our probabilities are p1= 0.6, 6 P2 = 0.3 and P3 = 0.1. The formula is a big one. It's n factorial divided by the product of all the k factorials times each probability raised to its corresponding k. So for our last question part B, the probabilities are 0.6 6 for coffee, 0.3 for tea, and 0.1 for juice with 10 total customers.
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What is the probability that exactly six ordered coffee, three ordered tea, and one ordered juice? Let's plug everything into our big formula. We need P of X coffee= 6, XT = 3, and X juice= 1. That's 10 factorial / 6 factorial * 3 factorial * 1 factorial. Then we multiply that by 0.6 * 0.3 * 0.1 to the 1st. Okay. The fraction part simplifies to 840. The probabilities become about 0.046 then
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0.026. 027 and 0.1. Multiply it all out and we get approximately 0.1058. Our final answer, 10.58%. Number 21. Daily temperature. The daily high temperature in a city varies around an average with most days near that average and fewer extremely hot or cool days. So for part A, what distribution describes these daily temperatures? Well, let's look at the clues. The data clusters around the average and it's symmetric. This is a classic. It's the normal
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distribution. This is the famous bell curve we use for continuous data that's clustered around a mean, which we call mu. Probabilities are found using the standard normal CDF denoted by that capital Greek letter ph. The formula is P of X is less than or equal to X equals PH of the quantity X - mu all / sigma. Okay, let's use it. Part B. Summer temperatures are normal with a mean of 78° F and a standard deviation of 5°. The question is, what percentage of days have
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temperatures below 70° F? We are going to walk through this calculation step by step. First, we need to calculate the Zcore for 70°. The formula for the Zcore is Z = X - mu / sigma. For us, that's 70 - 78, which is8 / 5. This gives us a zcore of -1.6. Now we just have to find the probability that Z is less than -1.6 by using a Z table. And the result is 0.0548 or 5.48%. Next up, problem 22. Quality control. A production line produces 4% defective items.
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We take a random sample of 10 items. Part A. Which distribution models the number of defective items. Let's see. We have a fixed number of trials and only two outcomes, defective or not? That's right. It's the binomial distribution. We have n equals 10 independent trials with a success probability p of 0.04. The probability of getting exactly k defective items is given by the formula p of x= k equals c of 10 comma k * 0.04 to the k power * 0.96 ^ of 10 - k. Okay,
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on to part b. We have a binomial distribution with n= 10 and p= 0.04. What is the probability that none of them are defective? Let's solve this. We need to calculate P of X equals 0. This means all 10 items have to be good. And the probability of an item being good is 96%. So using our formula P of X= 0 is C of 10, 0 * 0.04 04 to the power of 0 * 0.96 ^ of 10. Now C of 10 comma 0 is just 1 and 0.04 to the 0 is also 1. So it simplifies to just 0.96 to the 10th
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power which is approximately 0.6648. The result 66.48%. [Applause] Problem 23. Playing card. A single card is drawn from a well-shuffled standard deck of 52 cards. Part A. Which distribution models the suit hearts, diamonds, clubs, spades of the drawn card? Here's a hint. Is any suit more likely than another? Of course not. That's why it's a uniform distribution, the discrete version. Since there are 13 cards of each of the four suits, all outcomes are equally
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likely. The probability of X equaling any given suit is 13 divided by 52, which simplifies to 1/4. That's true for each of the four suits. So for part B, the probability of drawing any specific suit is 1/4. What is the probability the drawn card is either a heart or a diamond?
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This is a nice simple one. We want two specific suits out of four total suits. All we have to do is add their probabilities together. The probability of heart or diamond is the probability of a heart plus the probability of a diamond. That's just 1/4 + 1/4, which gives us 2/4s or 50%. Easy as that. Let's roll into problem 24. Rolling a specific number. You are rolling a standard six-sided die repeatedly until you get a four. Part A. Which distribution models the number
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of rolls until the first four? Notice we're not counting total successes here. We're waiting for the first one. This is the geometric distribution. It models the number of trials K required to get the very first success. For a success probability P of 16, the probability of the first four happening on roll K is P of X= K equals the quantity 1 - P to the power of K -1 * P. For us, that's 56 to the k minus one * 16. Let's apply this for part B. The probability of rolling a four
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is P = 16. What is the probability that the first time you roll a four is on your fifth roll? Okay, let's think this through. This requires four failures that is not a four followed by one success which is a four. We just plug k= 5 into our formula. p of x = 5 is the quantity 1 - 16 to the power of 5 - 1 * 16. That simplifies to 56 to the 4th power * 16. 56 to the 4th is about 0.4822. And we multiply that by 0.1667 giving us approximately 0.0804. The result 8.04%.
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Problem 25. Email arrivals. Emails arrive in your inbox at an average rate of 10 per hour. Part A. Which distribution models the time between two consecutive emails? If the arrivals are a pson process, what describes the waiting time? Hint, it's the memoryless one. You got it. It's the exponential distribution. This models the time between events in a pson process. The probability that the wait time x is less than or equal to some time t is given by the
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formula 1 minus e to the power of negative lambda t where lambda is the event rate. So for our final question part Bs arrive at an average rate of 10 per hour. What is the probability that the time between two consecutive emails is between 3 minutes and 9 minutes? Let's break it down. Step one, we have to convert our rate lambda to emails per minute. The rate is 10 emails per 60 minutes. So lambda is 16th of an email per minute. Super important step. Step two,
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we need to calculate P of 3 is less than X is less than 9. We do this by finding P of X is less than 9 and subtracting P of X is less than 3. Using our formula, that's the quantity 1 - e^ - 16 * 9us the quantity 1 - e - 16 * 3. Let's simplify those exponents. That becomes 1 - e^ -1.5 minus 1 - e to the0.5. When you distribute that negative sign, the ones cancel out and we're left with e to the0.5 minus e to the -1.5. Plugging in the values, that's about 0.6065
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minus 0.2231. 2231, which equals 0.3834. Our final result, 38.34%. Number 26. It's a classic, the coin flip. You flip a fair coin once. Part A. What distribution describes the outcome of getting heads? Think about it. We have one flip and only two possible outcomes. That's right. It's the Bernowi distribution. This is the perfect distribution for a single trial with two outcomes like success or failure. In this case, X= 1 for heads with
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a probability P of 0.5 and X= 0 for tails with a probability of 1 minus P which is also 0.5. Simple enough, right? So for part B, the outcome of a fair coin flip is a Berni trial with P equals 0.5. What is the probability of getting heads? And what is the variance? Well, the probability of heads is simply P. So that's an easy 50%. Now for the variance, the variance for a Bernoli distribution is P * the quantity 1 minus P. So we just plug in our numbers. 0.5 * 1 - 0.5, which
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equals 0.25. Easy peasy. Moving on to problem 27. Biological growth. The size of organisms often exhibits a rightke skewed distribution. Why? Because growth is proportional to current size. So part A, what distribution would you use to model such sizes? Here's a hint. When growth is multiplicative, taking the logarithm makes it normal. It's the lognormal distribution. A variable X is lognormally distributed if its natural logarithm lni is normally distributed.
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So if X is our organism size then the natural log of X follows a normal distribution with a mean of mu sublog and a variance of sigma sublog squared. Okay, let's apply this for part B. The log of a bacterium size that's ln of size is normal with a mean of 2.5 and a standard deviation of 0.3. The question is what is the probability that a bacterium is larger than 15 micrometers? First step one we have to transform our size threshold to the log scale. So we take
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the natural log of 15 which is approximately 2.708. [Music] Next step two calculate the zcore. Remember the formula it's our value minus the mean divided by the standard deviation. So we get z equ= 2.708us 2.5 all divided by 0.3. That works out to approximately 0.693. And finally, step three, we need to find the probability that Z is greater than 0.693 from a Z table. This is the same as 1 minus the probability that Z is less than or equal to 0.693.
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Looking that up, we get 1 minus 0.7557, giving us our final result, 24.43%. Problem 28. Typo count. A proofreader finds an average of two typos per page in a manuscript. Part A. Which distribution models the number of typos on a randomly chosen page? We're counting discrete events with a known average rate. Sound familiar? It's the Pson distribution. This distribution is perfect for modeling the number of events occurring in a fixed interval
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of time or space given a constant mean rate, which we call lambda. For lambda equals two typos per page, the probability of finding k typos is given by the formula you see here. lambda to the k * e to the negative lambda all divided by k factorial. So for part b, the rate of typos is lambda equals 2 per page. What is the probability that a randomly chosen page has exactly one typo? Let's use our formula. We need to calculate p of x= 1 using lambda= 2. So
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we plug in the numbers. P of X = 1 is 2 ^ of 1 * E -2 all divided by 1 factorial. That simplifies to just 2 * E -2 since E -2 is about 0.1353. We get 2 * that which is approximately 0.2707. The result 27.07%. Okay. Problem 29, medical trial. An experimental drug cures 60% of patients. If seven patients are treated, part A. Which distribution models the number of patients cured? We have a fixed
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number of patients and two outcomes for each cured or not cured. You know this one. It's the binomial distribution. We have n= 7 independent trials where each has a success probability of p= 0.6. The probability of exactly k patients being cured is given by the big formula here. C of 7 k * 0.6 ^ of k * 0.4 ^ of 7 minus k. Now for the application part B. We have a binomial distribution with n= 7 and p= 0.6. What is the probability that at least six are
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cured? At least six means we need to find the probability of exactly six being cured and the probability of exactly seven being cured and then add them together. So we need to calculate P of X is greater than or equal to 6 which equals P of X = 6 + P of X = 7. Let's do P of X= 6 first. That's C of 76 * 0.6 * 0.4 to the 1st. That gives us approximately 0.1306. Now for P of X= 7. That's C of 77 * 0.6 to the 7th * 0.4 to the 0 power. And that comes out to about 0.0279.
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Now we just add them up. 0.1306 plus 0.0279 gives us our result 15.85%.
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And our final problem number 30, lottery drawing. In a lottery, balls numbered 1 to 50 are placed in a machine and one ball is randomly selected. Part A. Which distribution models the number on the selected ball? Here's the key question. Does any ball have a better chance of being selected? Of course not. This is the uniform specifically the discrete uniform distribution. All 50 outcomes from ball number one to 50 are equally likely. The probability of selecting any specific ball
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K is simply 1 divided by 50. That leads us to our last question. Part B. There are 50 equally likely outcomes. What's the probability the selected ball has a number between 20 and 30 inclusive? Okay, let's think this through. The numbers we want are 20, 21, 22, all the way up to 30. How many numbers is that? Be careful here. It's not just 30 minus 20. It's 30 - 20 + 1, which gives us 11 favorable outcomes. The probability is just the ratio of those favorable outcomes
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to the total number of outcomes. So that's 11 divided by 50 and our result is 22%. [Music]
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