Question

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

inches and standard deviation 3.17 inches. Compute the probability that a simple random sample of size n=
10 results in a sample mean greater than 40 inches. That is, compute P(mean >40).
Gestation period The length of human pregnancies is approximately normally distributed with mean u = 266
days and standard deviation o = 16 days.
Tagged
Math
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days
or less?
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days
or less?
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of
the mean?
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Answer 1

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that
`\mu = 38.72, \sigma = 3.17`

Sample of 10:

This means that
`n = 10, s = (3.17)/(√(10))`

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (40 - 38.72)/((3.17)/(√(10)))`

`Z = 1.28`

`Z = 1.28` has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

`\mu = 266, \sigma = 16`

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

`Z = (X - \mu)/(\sigma)`

`Z = (260 -  266)/(16)`

`Z = -0.375`

`Z = -0.375` has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now
`n = 20`, so:

`Z = (X - \mu)/(s)`

`Z = (260 - 266)/((16)/(√(20)))`

`Z = -1.68`

`Z = -1.68` has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now
`n = 50`, so:

`Z = (X - \mu)/(s)`

`Z = (260 - 266)/((16)/(√(50)))`

`Z = -2.65`

`Z = -2.65` has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that
`n = 15`. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

`Z = (X - \mu)/(s)`

`Z = (276 - 266)/((16)/(√(15)))`

`Z = 2.42`

`Z = 2.42` has a p-value of 0.9922.

X = 256

`Z = (X - \mu)/(s)`

`Z = (256 - 266)/((16)/(√(15)))`

`Z = -2.42`

`Z = -2.42` has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.