Question

23. ITBS scores The Normal distribution with mean $\mu =6.8$and standard deviation $\sigma =1.6$is a good description of the Iowa Test of Basic Skills (ITBS) vocabulary scores of seventh-grade students in Gary, Indiana. Call the score of a randomly chosen student $X$for short. Find $P(X\ge 9)$and interpret the result. Follow the four-step process.

Short answer

The probability for $P(X\ge 9)$is $0.0838$. There is $8.38\%$chances that student's score is $9$or more.

Step-by-step solution

Given information

The Normal distribution with mean $\mu =6.8$and a standard deviation of $\sigma =1.6$.

Explanation

Let $X$be a random variable with a mean of $6.8$and a standard deviation of $1.6$and a normal distribution.

The$z-$score is calculated by dividing the value reduced by the mean by the standard deviation:
$z=\frac{x-\mu }{\sigma }$
Where, $x$indicates raw score, $\mu$indicates population mean, and$\sigma$ indicates the population standard deviation.
$P(X\ge 9)$can be calculated as:

$$\begin{gathered} P(x\ge 9)=P\left(\frac{x-\mu }{\sigma }\ge \frac{9-\mu }{\sigma }\right) \\ =P\left(z\ge \frac{9-6.8}{1.6}\right) \\ =P(Z\ge 1.38) \\ =0.0838 \end{gathered}$$