Chapter 6.8, Problem 14E

(a) .

To determine

To Calculate: The probability that the value of a random variable, having a normal distribution, is less than a certain value

Answer to Problem 14E

Explanation of Solution

Given Information:

• The mean of the normally distributed random variable is $\text{μ}=500g$ and the standard deviation is $\text{σ}=12g$
• The range for which the probability of occurrence is to be calculated is $0\le X\le 480$( the quantity of cereal in a box cannot be negative)

Concept Used:

• The probability distribution function of a normally distributed random variable with mean $\text{μ}$ and standard deviation $\text{σ}$ is given by
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$
• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$
• $e^{-x^2}$ does not have an anti-derivative. So, use a calculator or computer to calculate the definite integrals of such functions.

Calculation: Put $\text{a}=0$ and $\text{b}=480$ in the above integral for probability to write

  $\text{P}\left(0\le X\le 480\right)=\underset{0}{\overset{480}{{\displaystyle \int }}}\frac{1}{12\sqrt{2\text{π}}}{e}^{-\frac{{\left(x-500\right)}^2}{2\times {12}^2}}$

Input the values of lower limit $\text{a}$ , upper limit $\text{b}$ , the mean $\text{μ}$ and the standard deviation $\text{σ}$ in a calculator to get

  $\text{P}\left(0\le X\le 480\right)=0.0478=4.78\%$

Therefore, the probability that the machine produces a box with less than $480g$ of cereal is $0.0478$

(b).

To determine

To Calculate: The mean of a normal distribution when the standard deviation and the probability of a certain interval are given.

Answer to Problem 14E

The manufacturer should set a target weight of approximately $520g$ to be able to comply with the law.

Explanation of Solution

Given Information:

• The standard deviation is $\text{σ}=12$
• The probability that a box can have less than $500gm$ of cereal is $0.05$ , that is $5\%$ .

Concept Used:

• The probability distribution function of a normally distributed random variable with mean $\text{μ}$ and standard deviation $\text{σ}$ is given by
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$
• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$
• $e^{-x^2}$ does not have an anti-derivative. So, use a calculator or computer to calculate the definite integrals of such functions.

Calculation: Put $\text{a}=0$ and $\text{b}=500$ in the above integral for probability to write

  $P\left(0\le X<500\right)=\underset{0}{\overset{500}{{\displaystyle \int }}}\frac{1}{12\sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2\times {12}^2}}$

In the previous sub-part, a difference of $20$ between the mean ( $=500$ ) and the upper limit of the integral $b\left(=480\right)$ gave a probability of $4.78\%$ .

So, in the current sub-part too, the difference in the two numbers for a probability of $5\%$ should also be approximately the same value, that is $20$ .

Therefore,

  $\mu \approx b+20=500+20=520$

So, the manufacturer should set a target weight of approximately $520g$ to be able to comply with the law.