Chapter 13, Problem 18CT

a.

To determine

To find what percent of plants those are taller than $17$ inches.

Answer to Problem 18CT

  $0.62\%$ of plants are taller than $17$ inches.

Explanation of Solution

Given: Height of the plants are normally distributed with a mean of $12$ inches and a standard deviation of $2$ inches.

To find the z-scores use $z=\frac{x-\overline{x}}{\sigma }$ ,

For $x=17$ ,

  $\begin{array}{l}z=\frac{17-12}{2}\\ z=2.5\end{array}$

  $\begin{array}{l}P\left(x\ge 17\right)=P\left(z\le 2.5\right)\\ P\left(x\ge 17\right)=1-0.9938\to \text{using standard normal table}\\ P\left(x\ge 17\right)=0.0062\\ P\left(x\ge 17\right)=0.62\%\end{array}$

Thus we can find the percent of plants taller than $17$ inches.

b.

To determine

To find what percent of plants those are between $7\text{and}14$ inches.

Answer to Problem 18CT

  $83.51\%$ of plants are between $7\text{and}14$ inches.

Explanation of Solution

Given: Height of the plants are normally distributed with a mean of $12$ inches and a standard deviation of $2$ inches.

To find the z-scores use $z=\frac{x-\overline{x}}{\sigma }$ ,

For $x=7$ ,

  $\begin{array}{l}z=\frac{7-12}{2}\\ z=-2.5\end{array}$

For $x=14$ ,

  $\begin{array}{l}z=\frac{14-12}{2}\\ z=1\end{array}$

  $\begin{array}{l}P\left(7\le x\le 14\right)=P\left(-2.5\le z\le 1.0\right)\\ P\left(7\le x\le 14\right)=P\left(z\le 1.0\right)-P\left(z\le -2.5\right)\\ P\left(7\le x\le 14\right)=0.8413-0.0062\to \text{using standard normal table}\\ P\left(7\le x\le 14\right)=0.8351\\ P\left(7\le x\le 14\right)=83.51\%\end{array}$

Thus we can find the percent of plants are between $7\text{and}14$ inches.

c.

To determine

To find what percent of plants are at least $3$ inches taller than or $3$ inches lesser than the mean height.

Answer to Problem 18CT

  $13.36\%$ of plants are at least $3$ inches taller than or $3$ inches lesser than the mean height.

Explanation of Solution

Given: Height of the plants are normally distributed with a mean of $12$ inches and a standard deviation of $2$ inches.

To find the z-scores use $z=\frac{x-\overline{x}}{\sigma }$ ,

For $x=15\to \text{3 inch taller}$ ,

  $\begin{array}{l}z=\frac{15-12}{2}\\ z=1.5\end{array}$

For $x=9\to \text{3 inch taller}$ ,

  $\begin{array}{l}z=\frac{9-12}{2}\\ z=-1.5\end{array}$

  $\begin{array}{l}P\left(x\ge 17\text{or}x\le 9\right)=P\left(z\ge 1.5\text{or}z\le -1.5\right)\\ P\left(x\ge 17\text{or}x\le 9\right)=P\left(z\le -1.5\right)+\left(1-P\left(z\le 1.5\right)\right)\\ P\left(x\ge 17\text{or}x\le 9\right)=0.0668+\left(1-0.9332\right)\to \text{using standard normal table}\\ P\left(x\ge 17\text{or}x\le 9\right)=0.1336\\ P\left(x\ge 17\text{or}x\le 9\right)=13.36\%\end{array}$

Thus we can find the percent of plants are at least $3$ inches taller than or $3$ inches lesser than the mean height.