Mathematics, 10.03.2020 01:36 abbz13

For a certain type of computer, the length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours.
(a) John owns one of these computers and wants to know the probability that the length of time will be between 50 and 70 hours.
(b) What is the probability that the length of time between charges is more than 67.5?
(c) What is the probability that the length of time between charges is less than 48?
(d) What length of time between charges should a computer have to be in the top 5% of the curve?

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Answer from: starsinopoli13

a) 40.82% probability that the length of time will be between 50 and 70 hours.

b) 12.10% probability that the length of time between charges is more than 67.5 hours

c) 44.83% probability that the length of time between charges is less than 48 hours.

d) 74.675 hours

Step-by-step 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:

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

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:

$\mu = 50, \sigma = 15$

(a) John owns one of these computers and wants to know the probability that the length of time will be between 50 and 70 hours.

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 50. So

X = 70

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

$Z = \frac{70 - 50}{15}$

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

X = 50

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

$Z = \frac{50 - 50}{15}$

Z = 0

Z = 0 has a pvalue of 0.50

0.9082 - 0.50 = 0.4082

40.82% probability that the length of time will be between 50 and 70 hours.

(b) What is the probability that the length of time between charges is more than 67.5?

This is 1 subtracted by the pvalue of Z when X = 67.5. So

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

$Z = \frac{67.5 - 50}{15}$

Z = 1.17

Z = 1.17 has a pvalue of 0.8790

1 - 0.8790 = 0.1210

12.10% probability that the length of time between charges is more than 67.5 hours

pvalue of Z when X = 48. So

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

$Z = \frac{48 - 50}{15}$

Z = -0.13

Z = -0.13 has a pvalue of 0.4483

44.83% probability that the length of time between charges is less than 48 hours.

(d) What length of time between charges should a computer have to be in the top 5% of the curve?

Value of X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

$1.645 = \frac{X - 50}{15}$

X - 50 = 15*1.645

X = 74.675

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