Chapter 7.3, Problem 56E

a.

To determine

To find: The probability that randomly selected student would score 23 or more

Answer to Problem 56E

The probability is 0.3557.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population mean }\left(\mu \right)=21.1\\ \text{Population standard deviation}\left(\sigma \right)=5.1\end{array}$

Formula used:

The formula to compute the z -score is:

  $z=\frac{x-\mu }{\sigma }$

Where,

x is raw score

µ is population mean

s is population standard deviation.

Calculation:

Consider, X be the random variable follows the normal distribution with mean = 21.1 and standard deviation = 5.1.

The probability that randomly selected student would score 23 or more can be computed as:

  $\begin{array}{c}P\left(X>23\right)=P\left(\frac{x-\mu }{\sigma }>\frac{23-\mu }{\sigma }\right)\\ =P\left(Z>\frac{23-21.1}{5.0}\right)\\ =P\left(Z>0.37\right)\left(\text{From standard normal table}\right)\\ =0.3557\end{array}$

Thus, the required probability is 0.3557.

b.

To determine

To find: The probability that mean score of 50 students is 23 or above.

Answer to Problem 56E

The probability is 0.3974.

Explanation of Solution

Calculation:

The probability that mean score of 50 students is 23 or above is calculated as follows:

  $\begin{array}{c}P\left(\overline{X}>295\right)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{295-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ =P\left(Z>\frac{23-21.1}{\frac{5.1}{\sqrt{50}}}\right)\\ =P\left(Z>0.26\right)\left(\text{From standard normal table}\right)\\ =0.3974\end{array}$

Thus, the required probability is 0.3974.