Question

The time until recharge for a battery in a laptop computer under common conditions is normally distributed with a mean of 260 minutes and a standard deviation of 50 minutes. a. What is the probability that a battery lasts more than four hours? b. What are the quartiles (the 25% and 75% values) of battery life? c. What value of life in minutes is exceeded with 95% probability? 4.5.7 .WP . SS V

Answer 1

Answer:

There is a 65.54% probability that the battery lasts more than four hours.

The first quartile is 226.5 minutes.

The third quartile is 294 minutes.

A value of 131.5 minutes is excedeed with 95% probability.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

$\mu = 260, \sigma = 50$

a. What is the probability that a battery lasts more than four hours?

4 hours is 4*60 = 240 minutes. The pvalue of the Zscore of X = 240 is the probability that the battery lasts less than four hours. The subtraction of 100% by this value is the probability that the battery lasts more than four hours.

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{240 - 260}{50}$

$Z = -0.4$

A Zscore of -0.4 has a pvalue of 0.3446. This means that there is a 34.46% probability that the battery lasts less than four hours.

The probability that the battery lasts more than four hours is 1 - 0.3446 = 0.6554.

There is a 65.54% probability that the battery lasts more than four hours.

b. What are the quartiles (the 25% and 75% values) of battery life?

The first quartile is the value of X when Z has a pvalue of 0.25.

When the pvalue is 0.25, $Z = -0.67$. So:

$Z = \frac{X - \mu}{\sigma}$

$-0.67 = \frac{X- 260}{50}$

$X - 260 = -33.5$

$X = 226.5$

The first quartile is 226.5 minutes.

The third quartile is the value of X when Z has a pvalue of 0.75.

When the pvalue is 0.75, $Z = 0.68$. So:

$Z = \frac{X - \mu}{\sigma}$

$0.68 = \frac{X- 260}{50}$

$X - 260 = 34$

$X = 294$

The third quartile is 294 minutes.

What value of life in minutes is exceeded with 95% probability?

This is the value of X when the Zscore has a pvalue of 0.05. That is $Z = -2.57$

$Z = \frac{X - \mu}{\sigma}$

$-2.57 = \frac{X- 260}{50}$

$X - 260 = -128.5$

$X = 131.5$

A value of 131.5 minutes is excedeed with 95% probability.