Correcting for a Finite PopulationIn a study of babies born with very low birth weights, 275 children were given IQ tests at age 8, and their scores approximated a normal distribution with $\mu$= 95.5 and $\sigma$= 16.0 (based on data from “Neurobehavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm,” by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24). Fifty of those children are to be randomly selected without replacement for a follow-up study.

a.When considering the distribution of the mean IQ scores for samples of 50 children, should ${\sigma }_{\overline{x}}$be corrected by using the finite population correction factor? Why or why not? What is the value of ${\sigma }_{\overline{x}}$?

b.Find the probability that the mean IQ score of the follow-up sample is between 95 and 105.

The set of IQ scores of 275 children are normally distributed with a mean value equal to 95.5 and a standard deviation equal to 16.0.

A sample of size 50 is selected without replacement.

a.

If the sample is selected using the method of sampling without replacement, a correction factor needs to be applied to adjust the value of ${\sigma }_{\overline{x}}$.

The rule states that if the sample is selected without replacement, and the sample size is greater than 5% of the population size, the value of is multiplied by the following correction factor:

$CF=\sqrt{\frac{N-n}{N-1}}$,where

N is the population size, and

n is the sample size.

Here, a sample of size 50 is selected from a population of size 275.

Here, the method of sampling is without replacement.

Also,

$$\begin{gathered} 5\%\text{of}N=5\%\times 275 \\ =\frac{5}{100}\times 275 \\ =13.75 \end{gathered}$$

.

As the sample size (n=50) is greater than 5% of the population size (13.75), and the sampling is done without replacement, a correction factor needs to be multiplied to adjust the value of${\sigma }_{\overline{x}}$.

The corrected value of ${\sigma }_{\overline{x}}$is computed below.

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}CF \\ =\frac{\sigma }{\sqrt{n}}\sqrt{\frac{N-n}{N-1}} \\ =\frac{16}{\sqrt{50}}\sqrt{\frac{275-50}{275-1}} \\ =2.05 \end{gathered}$$

Thus, the value of ${\sigma }_{\overline{x}}$is equal to 2.05.

b.

Let $\overline{x}$denote the sample mean IQ score.

The sample mean IQ score follows a normal distribution with a mean equal to ${\mu }_{\overline{x}}=\mu$ and a standard deviation equal to ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The sample size is equal to n=50.

The probability that the sample mean IQ score is between 95 and 105 is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(95<\overline{x}<105\right)=P\left(\frac{95-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{105-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\right) \\ =P\left(\frac{95-95.5}{2.05}<z<\frac{105-95.5}{2.05}\right) \\ =P\left(-0.24<z<4.63\right) \\ =P\left(z<4.63\right)-P\left(z<-0.24\right) \end{gathered}$$

$$\begin{gathered} =0.9999-0.4052 \\ =0.5947 \end{gathered}$$

Therefore, the probability that the sample mean IQ score is between 95 and 105 is equal to 0.5947.