# Solved (Free): The time that it takes to assemble a piece of machinery is well modeled by the normal distribution with mean of

#### ByDr. Raju Chaudhari

Mar 9, 2021

The time that it takes to assemble a piece of machinery is well modeled by the normal distribution with mean of 72.9 minutes and standard deviation of 8.55 minutes.

(a) What is the probability that it will take less than an hour to assemble the next piece of machinery?
(b) What is the probability that it will take between 65 minutes and 75 minutes to assemble the next piece of machinery?
(c) What is the probability that it will take more than 80 minutes to assemble the next piece of machinery?
(d) Find the 16th percentile of the random variable, time taken to assemble. This is the time that is exceeded 84% of the time.

#### Solution

Let $X$ denote the time takes to assemble a piece of machinery.

Given that $\mu = 72.9$, $\sigma = 8.55$.

(a) The probability that it will take less than an hour to assemble the next piece of machinery, that is, the probability that $X$ is less than $60$ minutes is

 \begin{aligned} P(X < 60) & = P\bigg(\frac{X-\mu}{\sigma} < \frac{60-72.9}{8.55} \bigg)\\ & = P\big(Z < -1.51 \big)\\ &= 0.0657 \end{aligned}

(b) The probability that it will take between $65$ minutes and $75$ minutes to assemble the next piece of machinery is

 \begin{aligned} P(65 < X < 75) & = P(X < 75) -P(X < 65)\\ & = P\bigg(\frac{X- \mu}{\sigma} < \frac{75-72.9}{8.55} \bigg)- P\bigg(\frac{X-\mu}{\sigma} < \frac{65-72.9}{8.55} \bigg)\\ & = P\big(Z < 0.25 \big)-P\big(Z < -0.92 \big)\\ &= 0.597 -0.1777\\ &= 0.4193 \end{aligned}

(c) The probability that it will take more than $80$ minutes to assemble the next piece of machinery is

 \begin{aligned} P(X > 80) & = 1- P(X < 80)\\ & = 1- P\bigg(\frac{X-\mu}{\sigma} < \frac{80-72.9}{8.55} \bigg)\\ & = 1- P\big(Z < 0.83 \big)\\ &= 1- 0.7968\\ &= 0.2032 \end{aligned}

(d) Let the 16th percentile of the random variable, time taken to assemble is $a$.

 \begin{aligned} & P(X\leq a) =0.16\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{a-72.9}{8.55}\big)=0.16\\ &\Rightarrow P(Z< \frac{a-72.9}{8.55}\big)=0.16\\ &\Rightarrow \frac{a-72.9}{8.55}= -0.994\\ &\Rightarrow a = 72.9 + (-0.994)* 8.55\\ &\Rightarrow a = 64.4013 \end{aligned}