In example 2 the number of pins is discrete, how could you represent that using a density curve ?

Very good question! It turns out that, if Mike and Adam play a large number of games the distribution of their scores will be very well approximated by a normal distribution (even if their scores are discrete variables!). This is a consequence of something called the "Central Limit Theorem". Here is a video of Sal talking about it from the AP/ College Statistics series: https://www.khanacademy.org/math/ap-statistics/sampling-distribution-ap/what-is-sampling-distribution/v/central-limit-theorem

In Example 2: The hint says P(D < 0), why the probability of the difference between the two data has to be less than 0?

We have D = A - M. If D < 0, then it can happen only when M > A, which means Mike scores higher than Adam. P(D < 0) means probability of an event where Mike scores higher than Adam. Hope that helps.

Hiya Sal and everyone at Khan, thank you for all your hard work. It would be really nice if we could get a worked example of a probability of an absolute value as that is something that comes up in the practice questions but wasn't covered in the videos leading up to it. Like P(X |5|) or something like that.

P(X = |5|) = P(X = 5) I think you meant something different, since you can always just replace |5| with 5. If you meant to ask something like P(|X| > 5) this would be (I think) P(|X| > 5) = P(X < -5) + P(X > 5)

Main content

Course: Statistics and probability > Unit 9

Lesson 4: Combining random variables

Combining normal random variables

Q: In Example 2: The hint says P(D < 0), why the probability of the difference between the two data has to be less than 0?

We have D = A - M. If D < 0, then it can happen only when M > A, which means Mike scores higher than Adam. P(D < 0) means probability of an event where Mike scores higher than Adam. Hope that helps.

Q: Hiya Sal and everyone at Khan, thank you for all your hard work. It would be really nice if we could get a worked example of a probability of an absolute value as that is something that comes up in the practice questions but wasn't covered in the videos leading up to it. Like P(X |5|) or something like that.

P(X = |5|) = P(X = 5) I think you meant something different, since you can always just replace |5| with 5. If you meant to ask something like P(|X| > 5) this would be (I think) P(|X| > 5) = P(X < -5) + P(X > 5)

Google Classroom

When we combine variables that each follow a normal distribution, the resulting distribution is also normally distributed. This lets us answer interesting questions about the resulting distribution.

Example 1: Total amount of candy

Each bag of candy is filled at a factory by

4

machines. The first machine fills the bag with blue candies, the second with green candies, the third with red candies, and the fourth with yellow candies. The amount of candy each machine dispenses is normally distributed with a mean of

50 g

and a standard deviation of

5 g

. Also, assume that the amount dispensed by any given machine is independent from the other machines.

Let

T

be the total weight of candy in a randomly selected bag.

Find the probability that a randomly selected bag contains less than $178 g$ ‍ of candy.

Let's solve this problem by breaking it into smaller pieces.

Problem A (Example 1)

Find the mean of $T$ ‍.

μ_{T} =

grams

Problem B (Example 1)

Find the standard deviation of $T$ ‍.

σ_{T} =

grams

Problem C (Example 1)

What shape does the distribution of $T$ ‍ have?

Problem D (Example 1)

Find the probability that a randomly selected bag contains less than $178 g$ ‍ of candy.
Round to four decimal places.

P (T < 178 g) \approx

Here is a partial standard normal table whose entries represent area below a given

z

-score.

Example 2: Difference in bowling scores

Adam and Mike go bowling every week. Adam's scores are normally distributed with a mean of

175

pins and a standard deviation of

30

pins. Mike's scores are normally distributed with a mean of

150

pins and a standard deviation of

40

pins. Assume that their scores in any given game are independent.

Let

A

be Adam's score in a random game,

M

be Mike's score in a random game, and

D

be the difference between Adam's and Mike's scores where

D = A - M

Find the probability that Mike scores higher than Adam in a randomly selected game.

Let's solve this problem by breaking it into smaller pieces.

Problem A (Example 2)

Find the mean of $D$ ‍.

μ_{D} =

pins

Problem B (Example 2)

Find the standard deviation of $D$ ‍.

σ_{D} =

pins

Problem C (Example 2)

What shape does the distribution of $D$ ‍ have?

Problem D (Example 2)

Find the probability that Mike scores higher than Adam in a randomly selected game.
Round to four decimal places.

P (Mike scores higher) \approx

Hint: Find $P (D < 0)$ ‍.

Here is a partial standard normal table whose entries represent area below a given

z

-score.

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MrJorge
Posted 6 years ago. Direct link to MrJorge's post “In Example 2: The hint sa...”
In Example 2: The hint says P(D < 0), why the probability of the difference between the two data has to be less than 0?
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(6 votes)
Answer
- Angels Angels
  Posted 6 years ago. Direct link to Angels Angels's post “We have D = A - M. If D <...”
  We have D = A - M. If D < 0, then it can happen only when M > A, which means Mike scores higher than Adam.
  P(D < 0) means probability of an event where Mike scores higher than Adam.
  Hope that helps.
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  (28 votes)
Mohamed Ibrahim
Posted 4 years ago. Direct link to Mohamed Ibrahim's post “In example 2 the number o...”
In example 2 the number of pins is discrete, how could you represent that using a density curve ?
Button navigates to signup pageComment on Mohamed Ibrahim's post “In example 2 the number o...”
(7 votes)
Answer
- Nic
  Posted 10 months ago. Direct link to Nic's post “Very good question! It tu...”
  Very good question! It turns out that, if Mike and Adam play a large number of games the distribution of their scores will be very well approximated by a normal distribution (even if their scores are discrete variables!). This is a consequence of something called the "Central Limit Theorem". Here is a video of Sal talking about it from the AP/ College Statistics series: https://www.khanacademy.org/math/ap-statistics/sampling-distribution-ap/what-is-sampling-distribution/v/central-limit-theorem
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  (3 votes)
owen-k
Posted 6 years ago. Direct link to owen-k's post “In the Practice quiz they...”
In the Practice quiz they keep having an absolute value probability question. How does one go about solving that? For instance one example is Sam's mean of washing cars is 20 minutes with a standard deviation of 6.4 minutes. Taylor's mean of washing the interior of cars is 18 minutes with a mean of 4.8 minutes.
Then it says find the probability that a randomly selected time of Sam and Taylor falls within 10 minutes of each other and gives the equation find P(D less than |10|). So I do D=(S-T) and I get mean of D is 2 minutes and the standard deviation is 8 minutes. So far so good, but after that I always go wrong somehow. When I click on the explanation it says to do two z scores one of -10 and one of 10 and then calculate between them, but why would I do -10 and 10, it says within ten minutes of each other, wouldn't that mean you would do ten above and ten below the mean of D?
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(6 votes)
Answer
- Brian.Wentroble
  Posted 4 years ago. Direct link to Brian.Wentroble's post “The mean and standard dev...”
  The mean and standard deviation explain the shape of the curve and can tell which percentages are above and below certain points. However, the question asks whether they finish within 10 minutes of each other. Since Taylor is 2 minutes quicker than Sam, the area under the curve is shifted. The center where they both have the same time is 0, reflecting where both Sam and Taylor have the same finish time (S-T). Calculate 10 minutes below and 10 minutes above 0, the place where they are equal, to find the percentages where they are finishing within 10 minutes of each other.
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  (1 vote)
alivia.ryan
Posted 4 months ago. Direct link to alivia.ryan's post “Im confused on how you do...”
Im confused on how you do the last part (problem D) for both examples
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(1 vote)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “In problem D of both exam...”
  In problem D of both examples, you're essentially finding the probability of a specific event occurring based on the normal distribution. This involves calculating the z-score corresponding to the given value (e.g., less than 180 grams of candy or Mike scoring higher than Adam in bowling) and then using a standard normal distribution table or calculator to find the corresponding probability.
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  (1 vote)
allegra.vanklink
Posted 2 months ago. Direct link to allegra.vanklink's post “In example one, you give ...”
In example one, you give the z score for -2.2 as 0.0139, but on every other z table I look at, it's 0.41294. What's going on there?
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(1 vote)
Answer
- daniella
  Posted a month ago. Direct link to daniella's post “The discrepancy in the z-...”
  The discrepancy in the z-score and resulting probability might be due to different methods of calculation or rounding. It's essential to ensure consistency in the method of calculation and rounding when comparing results from different sources. If you're using a standard normal distribution table or calculator, the value of P(D < −2.2) should indeed be closer to 0.0139. If you're getting a different value, it's worth double-checking your calculation or the source of the z-table you're using.
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  (1 vote)
owen-k
Posted 6 years ago. Direct link to owen-k's post “Hiya Sal and everyone at ...”
Hiya Sal and everyone at Khan, thank you for all your hard work. It would be really nice if we could get a worked example of a probability of an absolute value as that is something that comes up in the practice questions but wasn't covered in the videos leading up to it. Like P(X |5|) or something like that.
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(0 votes)
Answer
- Bryan
  Posted 4 years ago. Direct link to Bryan's post “P(X = |5|) = P(X = 5) I ...”
  P(X = |5|) = P(X = 5)
  
  I think you meant something different, since you can always just replace |5| with 5.
  
  If you meant to ask something like
  P(|X| > 5)
  this would be (I think)
  P(|X| > 5) = P(X < -5) + P(X > 5)
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  (2 votes)
David Schamberger
Posted 6 years ago. Direct link to David Schamberger's post “For example 2, I am confu...”
For example 2, I am confused why we are finding P(D<0). Since, Adam and Mike are playing bowling the difference of the two normal distributions must be discrete whole numbers? Then, D can't be any value between -1 and 0. Also, 0 isn't part of the solution space. Then we should be trying to find P(D<=-1)?

<= means less than or equal too.
Button navigates to signup pageComment on David Schamberger's post “For example 2, I am confu...”
(1 vote)
Answer
- Anant Sadhu
  Posted 6 years ago. Direct link to Anant Sadhu's post “As r.v. D = A - M. The on...”
  As r.v. D = A - M. The only possible way to that Mike has more pins than Adam is when Mike's pins is greater than Adam's.
  Therefore, for our question D should be a negative number. That's why in hint D < 0
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  (0 votes)
Kaciyah Rook
Posted 5 years ago. Direct link to Kaciyah Rook's post “what happens when the pro...”
what happens when the probability is greater than a set mean? for example P(X>14)
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Answer
- daniella
  Posted a month ago. Direct link to daniella's post “When the probability is g...”
  When the probability is greater than a set mean, for example P(X > 14), it means you're calculating the probability of the random variable being greater than 14. This can be interpreted as finding the probability of an event occurring where the outcome is greater than 14.
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  (1 vote)
alaatamimi294
Posted 5 years ago. Direct link to alaatamimi294's post “again makes more sense th...”
again makes more sense that D=M-A, since the probability in demand is M scoring more than A
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Answer
- daniella
  Posted a month ago. Direct link to daniella's post “In the context of Example...”
  In the context of Example 2, if D = M − A, it would represent the difference in scores where Mike scores more than Adam. However, the problem defines D = A − M, representing the difference in scores where Adam scores more than Mike. Both formulations are valid, but the problem explicitly defines the order of subtraction.
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  (1 vote)