Chapter 7.2, Problem 9E

To determine

(a)

To find:

Find the probability of a random person on the street having an IQ score of less than 95.

Answer to Problem 9E

Solution:

The probability of a random person on the street having an IQ score of less than 95 is 0.3707.

Explanation of Solution

Given:

$\mu =100and\sigma =15$

Given a population of size $N$ with mean $\left(\mu \right)$ and standard deviation $\left(\sigma \right)$, samples of a fixed discrete number of trials $\left(n\right)$ can be chosen each having a mean called the sample mean $\left(\overline{x}\right)$.

On dealing with the collection of these means for samples chosen from the population, the concept of sampling distribution pops and the statistic dealt with is the sample means, and that the distribution contains all possible samples for the chosen sample size.

The continuity criterion can be extended and here it is known as Central Limit Theorem which states that sampling distribution is approximately normal with:

a. mean ${\mu }_{\overline{x}}=\mu$, and

b. Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Moreover, the standard score is given by:

$z=\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}$.

Thereafter, the required probability is accordingly obtained by taking into account the specifics in the question.

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{95-100}{15}\\ & =& \frac{-5}{15}\\ & =& -0.333\end{array}$

Now probability of x less than 95 is calculated as

$\begin{array}{rll}P\left(x<95\right)& =& P\left(z<-0.33\right)\\ & =& 0.3707\end{array}$

So, the probability of a random person on the street having an IQ score of less than 95 is 0.3707.

To determine

(b)

To find:

Find the probability that the mean of the sample taken is less than 95.

Answer to Problem 9E

Solution:

The probability that the mean of the sample taken is less than 95 is 0.0091.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{95-100}{\left(\frac{15}{\sqrt{50}}\right)}\\ & =& \frac{-5}{2.12}\\ & =& -2.36\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}<95\right)& =& P\left(z<-2.36\right)\\ & =& 0.0091\end{array}$

So the probability that the mean of the sample taken is less than 95 is 0.0091.

To determine

(c)

To find:

Find the probability that the mean of the sample taken will be more than 95.

Answer to Problem 9E

Solution:

The probability that the mean of the sample taken will be more than 95 is 0.9909.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{95-100}{\left(\frac{15}{\sqrt{50}}\right)}\\ & =& \frac{-5}{2.12}\\ & =& -2.36\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}>95\right)& =& P\left(z>-2.36\right)\\ & =& 1-P\left(z<-2.36\right)\\ & =& 1-0.0091\\ & =& 0.9909\end{array}$

The probability that the mean of the sample taken will be more than 95 is 0.9909.

To determine

(d)

To find:

Find the probability that the mean of the sample taken will be more than 105.

Answer to Problem 9E

Solution:

The probability that the mean of the sample taken will be more than 105 is 0.0091.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{105-100}{\left(\frac{15}{\sqrt{50}}\right)}\\ & =& \frac{5}{2.12}\\ & =& 2.36\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}>105\right)& =& P\left(z>2.36\right)\\ & =& 1-P\left(z<2.36\right)\\ & =& 1-0.9909\\ & =& 0.0091\end{array}$

The probability that the mean of the sample taken will be more than 105 is 0.0091.

To determine

(e)

To find:

Find the probability that the sample mean differs from the population mean by more than 5.

Answer to Problem 9E

Solution:

The probability that the sample mean differs from the population mean by more than 5 is 0.0182.

Explanation of Solution

Calculation:

The probability that the sample mean differs from the population mean by more than 5 is calculated as:

$P\left(\left|\overline{x}-\mu \right|\right)>5$

$P\left(\left|\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\right|\right)>\frac{5}{\frac{15}{\sqrt{50}}}$

$$ P\left(\left|z\right|\right)>\frac{5}{2.12}$$

$$ P\left(z\right)>2.36;P\left(z\right)<-2.36$$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\left|z\right|>2.36\right)& =& P\left(z<-2.36\right)+P\left(z>2.36\right)\\ & =& P\left(z<-2.36\right)+1-P\left(z<2.36\right)\\ & =& 0.0091+1-0.9909\\ & =& 0.0182\end{array}$