Chapter 7, Problem 5RE

Lightbulbs: The lifetime of lightbulbs has a mean of 1500 hours and a standard deviation of 100 hours. A sample of 50 lightbulbs is tested.

1. What is the probability that the sample mean lifetime is greater than 1520 hours?
2. What is the probability that the sample mem lifetime is less than 1540 hours?
3. What is the probability that the sample mem lifetime is between 1490 and 1550 hours?

To determine

To find: the probability that the sample mean lifetime is greater than $1520$ hours.

Answer to Problem 5RE

The required answer is $0.079$ .

Explanation of Solution

Given Information:

Given that the population mean is $\mu =1500$

standard deviation is $\sigma =100$ .

  $n=50$ .

Formula used:

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Required Calculations:

The standard deviation is,

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{100}{\sqrt{50}}\\ =14.14\end{array}$

It is asked to find $P\left(\overline{X}>1520\right)$

  $\begin{array}{l}P\left(\overline{X}>1520\right)=P\left(\frac{X-1500}{14.4}>\frac{1520-1500}{14.4}\right)\\ P\left(\overline{X}>1520\right)=P\left(Z>1.388\right)\end{array}$

Using normal tables

  $P\left(\overline{X}>1520\right)=0.079$

Calculation:

The standard deviation is,

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{100}{\sqrt{50}}\\ =14.14\end{array}$

To compute the probability, that $\overline{x}$ will be greater than $1520$ hours using Minitab follow the steps given below.

1. Click CaIc, then select Probability distributions and then go to Normal.

2. Select the Cumulative option.

3. Enter $1500$ in the Mean field and enter $14.14$ in the standard deviation field.

4. Enter $1520$ in the input constant field.

5. Click Ok. The final output is given below.

Cumulative Distribution function

Normal with mean $=1500$ and standard deviation $=14.14$

  $\begin{array}{l}xP\left(X\le x\right)\\ 15200.921382\end{array}$

The required probability that $\overline{x}$ is greater than $1520$ is,

  $\begin{array}{l}P\left(\overline{x}>1520\right)\\ =1-P\left(\overline{x}<1520\right)\\ =1-0.921382\\ =0.079\end{array}$

The required answer is $0.079$ .

To determine

To find: the probability that the sample mean lifetime is less than 1540 hours.

Answer to Problem 5RE

The required probability is 0.997664.

Explanation of Solution

Given Information:

Given that the population mean is $\mu =1500$

standard deviation is $\sigma =100$ .

  $n=50$ .

Required Calculations:

The standard deviation is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{100}{\sqrt{50}}\\ =14.14\end{array}$

To find: $P\left(\overline{X}<1540\right)$

  $\begin{array}{l}P\left(\overline{X}<1540\right)=P\left(\frac{\overline{X}-1540}{14.4}>\frac{1540-1500}{14.4}\right)\\ P\left(\overline{X}<1540\right)=P\left(Z<2.77\right)\end{array}$

Using normal table

  $P\left(\overline{X}<1540\right)=0.997664$

To determine

To find: the probability that the sample mean lifetime is between 1490 and 1550 hours.

Answer to Problem 5RE

The required probability is 0.76.

Explanation of Solution

Given Information:

Given that the population mean is $\mu =1500$

standard deviation is $\sigma =100$ .

  $n=50$ .

Calculation:

The standard deviation is,

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{100}{\sqrt{50}}\\ =14.14\end{array}$

To find: $P\left(1490<\overline{X}<1550\right)$

  $\begin{array}{l}P\left(1490<\overline{X}<1550\right)=P\left(\frac{1490-1500}{14.4}<\frac{\overline{X}-1540}{14.4}<\frac{1550-1500}{14.4}\right)\\ P\left(1490<\overline{X}<1550\right)=P\left(0.694<Z<3.472\right)\end{array}$

Using normal tables$P\left(1490<\overline{X}<1550\right)=0.76$