Chapter 6.1, Problem 23E

To determine

To find: $P\left(X\ge 9\right)$ and interpret it.

Answer to Problem 23E

The probability is 0.0838

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population mean }\left(\mu \right)=6.8\\ \text{Population standard deviation}\left(\sigma \right)=1.6\end{array}$

Formula used:

The formula to compute the z -score is:

  $z=\frac{x-\mu }{\sigma }$

Where,

x is raw score

µ is population mean

σ is population standard deviation.

Calculation:

Consider, X be the random variable follows the normal distribution with mean = 6.8 and standard deviation = 1.6

  $P\left(X\ge 9\right)$ can be computed as:

  $\begin{array}{c}P\left(X\ge 9\right)=P\left(\frac{x-\mu }{\sigma }\ge \frac{9-\mu }{\sigma }\right)\\ =P\left(Z\ge \frac{9-6.8}{1.6}\right)\\ =P\left(Z\ge 1.38\right)\left(\text{From standard normal table}\right)\\ =0.0838\end{array}$

Thus, the required probability is 0.0838.

Interpretation:

There are 8.38% chances that student’s score is 9 or more.