Question

Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by greater than 42 grams? Round your answer to four decimal places.

Answer 1

`P =0.3998`

Explanation:

Let
`{\displaystyle {\overline{x}}}` be the average of the sample, and the population mean will be
`\mu`

We know that:

`\mu = 4095` gr

Let
`\sigma` be the standard deviation and n the sample size, then we know that the standard error of the sample is:

`E=(\sigma)/(√(n))`

Where

`\sigma=569`

`n=130`

In this case we are looking for:

`P(|{\displaystyle{\overline{x}}}- \mu|>42)`

This is:

`{\displaystyle{\overline{x}}}- \mu>42` or
`{\displaystyle{\overline{x}}}- \mu<-42`

`P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu<-42)`

Now we get the z score

`Z=\frac{{\displaystyle{\overline{x}}}-\mu}{(\sigma)/(√(n))}`

`P=P(z>(42)/((569)/(√(130)))) + P(z<-(42)/((569)/(√(130))))`

`P=P(z>0.8416) + P(z<-0.8416)`

Looking at the tables for the standard nominal distribution we get

`P =0.1999+0.1999`

`P =0.3998`