Question

IQ tests The Wechsler Adult Intelligence Scale (WAIS) is a common IQ test for adults. The distribution of WAIS scores for persons over 16 years of age is approximately Normal with mean 100 and standard deviation $15 .

a. What is the probability that a randomly chosen individual has a WAIS score of 105 or greater?

b. Find the mean and standard deviation of the sampling distribution of the average WAIS score ${\mathrm{x}}^{-}\overline{x}$for an SRS of 60 people. Interpret the standard deviation.

c. What is the probability that the average WAIS score of an SRS of 60 people is 105 or greater?

d. Would your answers to any of parts (a), (b), or (c) be affected if the distribution of WAIS scores in the adult population was distinctly non-Normal? Explain your reasoning.

Short answer

(a)The associatively probability using table =0.3707

(b)The mean of the sampling distribution of the sample mean $\overline{x}$is

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{15}{\sqrt{60}}=1.9365$

(c)The corresponding probability using table

role="math" localid="1654404119865" $$\begin{gathered} P(\overline{x}\ge 105)=P(z>2.58) \\ =P(z<-2.58) \\ =0.0049 \end{gathered}$$

(d) the sampling distribution is about normal even if the population distribution is not normal.

Step-by-step solution

Part (a) Step 1: Given Information

$$\begin{gathered} \mu =100 \\ \sigma =15 \end{gathered}$$

Formula used:

$z=\frac{x-\mu }{\sigma }$

Part (a) Step 2: Simplification

The z-score is

$$\begin{gathered} =\frac{x-\mu }{\sigma } \\ =\frac{105-100}{15} \\ =0.33 \end{gathered}$$

The associatively probability using table

$$\begin{gathered} P(X\ge 105)=P(z>0.33) \\ =P(z<-0.33) \\ =0.3707 \end{gathered}$$

Part (b) Step 1: Given Information

$$\begin{gathered} \backslash mu\&=100\backslash \backslash \\ \backslash sigma\&=15\backslash \backslash \\ n\&=60 \end{gathered}$$

Formula used:

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Part (b) Step 2: Simplification

The mean of the sampling distribution of the sample mean $\overline{x}$is

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{15}{\sqrt{60}}=1.9365$

Part (c) Step 1: Given Information

$$\begin{gathered} \mu =100 \\ \sigma =15 \\ n=60 \end{gathered}$$

Formula used:

$$\begin{gathered} {\sigma }_x=\frac{\sigma }{\sqrt{n}} \\ z=\frac{x-\mu }{\sigma } \end{gathered}$$

Part (c) Step 2: Simplification

The mean of the sampling distribution of the sample mean $\overline{x}$is same to the population standard deviation divided by the square root of the sample size

${\mu }_{\overline{x}}=\mu =100$

The standard deviation of the sampling distribution of the sample mean $\overline{x}$is

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{15}{\sqrt{60}}=1.9365$

The z-score is

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{105-100}{1.9365} \\ =2.58 \end{gathered}$$

The corresponding probability using table

$$\begin{gathered} P(\overline{x}\ge 105)=P(z>2.58) \\ =P(z<-2.58) \\ =0.0049 \end{gathered}$$

Part (d) Step 1: Simplification

The answer in (a) could be very different, because it is supposed that the population distribution was normal to find the probability.

The answer in (b) and (c) will not be very different, the reason is that the sample size is 30 or more and, by the central limit theorem, know that the sampling distribution is about normal even if the population distribution is not normal.