91影视

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 ${蟽}_{\stackrel{炉}{x}}$.

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,

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

.

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${蟽}_{\stackrel{炉}{x}}$.

The corrected value of ${蟽}_{\stackrel{炉}{x}}$is computed below.

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

Thus, the value of ${蟽}_{\stackrel{炉}{x}}$is equal to 2.05.

b.

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

The sample mean IQ score follows a normal distribution with a mean equal to ${渭}_{\stackrel{炉}{x}}=渭$ and a standard deviation equal to ${蟽}_{\stackrel{炉}{x}}=\frac{蟽}{\sqrt{n}}$.

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.

$$\begin{gathered} P\left(95<\stackrel{炉}{x}<105\right)=P\left(\frac{95-{渭}_{\stackrel{炉}{x}}}{{蟽}_{\stackrel{炉}{x}}}<\frac{\stackrel{炉}{x}-{渭}_{\stackrel{炉}{x}}}{{蟽}_{\stackrel{炉}{x}}}<\frac{105-{渭}_{\stackrel{炉}{x}}}{{蟽}_{\stackrel{炉}{x}}}\right) \\ =P\left(\frac{95-95.5}{2.05}<z<\frac{105-95.5}{2.05}\right) \\ =P\left(-0.24<z<4.63\right) \\ =P\left(z<4.63\right)-P\left(z<-0.24\right) \end{gathered}$$

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

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