Question

Assume that the duration of human pregnancy can be described by a Normal model with mean 266 days and standard deviation 16 days.

1) What is the probability that the duration of the pregnancy of a randomly chosen pregnant woman will last less than 260 days?
Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Consider the sample mean duration of the pregnancies of the 60 pregnant women.
2) What is the mean of the sampling distribution of the sample mean?
3) What is the standard deviation of the sampling distribution of the sample mean?
4) What is the model for the sampling distribution?
5) What is the probability that the sample mean duration of the 60 pregnant women will be less than 260 days? Explain why this answer is different from the answer to part 36.

Answer 1

1)
`P(X<260)=P((X-\mu)/(\sigma)<(260-\mu)/(\sigma))=P(Z<(260-266)/(16))=P(Z<-0.375)`

And we can find this probability using the normal standar ddistirbution or excel:

`P(Z<-0.375)=P(Z<-0.375)=0.354`

2)
`\mu_(\bar X) = \mu=266`

3)
`\sigma_(\bar x)= (\sigma)/(√(n))= (16)/(√(60))=2.07`

4)
`\bar X \sim N(\mu=266, \sigma_(\bar X)= (\sigma)/(√(n))=2.07)`

5)
`P(\bar X <260)=P(Z<(260-266)/((16)/(√(60)))=-2.905)`

And using a calculator, excel ir the normal standard table we have that:

`P(Z<-2.905)=0.00183`

This probability is different because for this case we are finding the probability that the sample mean with 60 observations would be less than 260, and from part 1 we were finding an individual probability.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part 1

Let X the random variable that represent the duration of the pregnancy of a population, and for this case we know the distribution for X is given by:

`X \sim N(266,16)`

Where
`\mu=266` and
`\sigma=16`

We are interested on this probability

`P(X<260)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

`P(X<260)=P((X-\mu)/(\sigma)<(260-\mu)/(\sigma))=P(Z<(260-266)/(16))=P(Z<-0.375)`

And we can find this probability using the normal standar distribution or excel:

`P(Z<-0.375)=P(Z<-0.375)=0.354`

Part 2

For this case the mean of the sampling distribution is:

`\mu_(\bar X) = \mu=266`

Part 3

The standard deviation for the sample mean is given by:

`\sigma_(\bar x)= (\sigma)/(√(n))= (16)/(√(60))=2.07`

Part 4

The distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu=266, \sigma_(\bar X)= (\sigma)/(√(n))=2.07)`

Part 5

We can find the probabilitiy like this:

`P(\bar X <260)=P(Z<(260-266)/((16)/(√(60)))=-2.905)`

And using a calculator, excel ir the normal standard table we have that:

`P(Z<-2.905)=0.00183`

This probability is different because for this case we are finding the probability that the sample mean with 60 observations would be less than 260, and from part 1 we were finding an individual probability.