Question

The time until recharge for a battery in a laptop computer under common conditions is normally distributed with mean of 265 minutes and a standard deviation of 50 minutes. a) What is the probability that a battery lasts more than four hours? Enter your answer in accordance to the item a) of the question statement (Round the answer to 3 decimal places.) b) What are the quartiles (the 25% and 75% values) of battery life? 25% value = Enter your answer; 25% value = _ minutes minutes (Round the answer to the nearest integer.) 75% value = Enter your answer; 75% value = _ minutes minutes (Round the answer to the nearest integer.) c) What value of life in minutes is exceeded with 95% probability? Enter your answer in accordance to the item c) of the question statement (Round the answer to the nearest integer.)

Answer 1

a) 0.691 = 69.1% probability that a battery lasts more than four hours

b) 25% value = 231

75% value = 299

c) 183 minutes

Explanation:

Problems of normally distributed samples are 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 = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 265, \sigma = 50`

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

4 hours = 4*60 = 240 minutes

This is 1 subtracted by the pvalue of Z when X = 240. So

`Z = (X - \mu)/(\sigma)`

`Z = (240 - 265)/(50)`

`Z = -0.5`

`Z = -0.5` has a pvalue of 0.309

1 - 0.309 = 0.691

0.691 = 69.1% probability that a battery lasts more than four hours

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

25th percentile:

X when Z has a pvalue of 0.25. So X when Z = -0.675

`Z = (X - \mu)/(\sigma)`

`-0.675 = (X - 265)/(50)`

`X - 265 = -0.675*50`

`X = 231`

75th percentile:

X when Z has a pvalue of 0.75. So X when Z = 0.675

`Z = (X - \mu)/(\sigma)`

`0.675 = (X - 265)/(50)`

`X - 265 = 0.675*50`

`X = 299`

25% value = 231

75% value = 299

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

The 100-95 = 5th percentile, which is the value of X when Z has a pvalue of 0.05. So X when Z = -1.645.

`Z = (X - \mu)/(\sigma)`

`-1.645 = (X - 265)/(50)`

`X - 265 = -1.645*50`

`X = 183`