Chapter 11.1, Problem 17E

To determine

To find: the probability where $x-value$ is in the given interval.

Answer to Problem 17E

The probability that a randomly selected $x-value$ lies below $37$ is $0.84$ .

Explanation of Solution

Given:

Mean is $33$

Standard deviation is $4$ .

At most $37$ .

Calculation:

In this question, normal distribution has a mean of $33$ and standard deviation of $4$ .

That is $\mu =33$ and $\sigma =4$ . Thus,

  $\begin{array}{l}\mu -3\sigma =33-3\left(4\right)=21\\ \mu -2\sigma =33-2\left(4\right)=25\\ \mu -\sigma =33-4=29\\ \mu +\sigma =33+4=37\\ \mu +2\sigma =33+2\left(4\right)=41\\ \mu +3\sigma =33+3\left(4\right)=45\end{array}$

Thus the graph of this normal distribution looks like:

  

To find area under normal curve, these properties of a normal distribution is used,

The total area under the normal curve is $1$ .

About $68\%$ of the area lies within $1$ standard deviation of mean.

About $95\%$ of the area lies within $2$ standard deviation of mean.

About $99.7\%$ of the area lies within $3$ standard deviation of mean.

  

The value $37$ represents $\mu +\sigma$ that is one standard deviation above the mean.

Now, comparing the first and second diagrams, the probability that a randomly selected $x-value$ lies below $37$ is given by

  $\begin{array}{l}P\left(x\le 37\right)=0.34+0.34+0.135+0.0235+0.0015\\ =0.84\end{array}$

Conclusion:

Hence, the probability that a randomly selected $x-value$ lies below $37$ is $0.84$ .