Question

A manufacturer claims that the batteries it makes will last 18 hours, with a standard deviation of 1.5 hours. If the durations of the batteries are normally distributed, what proportion of batteries would be expected to last less than 16 hours?

A. 0.9082

B. 0.0918

C. 0.1134

D. 0.2537

E. 0.5918

Answer 1

Answer: the correct option is B

Explanation:

the durations of the batteries are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = durations of the batteries in hours

u = mean time

s = standard deviation

From the information given,

u = 18 hours

s = 1.5 hours

We want to find the proportion or probability of batteries would be expected to last less than 16 hours. It is expressed as

P(x lesser than 16)

For x = 16,

z = (16 - 18)/1.5 = - 1.33

Looking at the normal distribution table, the probability corresponding to the z score is 0.09176

Approximately 0.0918

Answer 2

B. 0.0918

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 = (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 = 18, \sigma = 1.5`

What proportion of batteries would be expected to last less than 16 hours?

This is the pvalue of Z when X = 16. So:

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

`Z = (16 - 18)/(1.5)`

`Z = -1.33`

`Z = -1.33` has a pvalue of 0.0918.

So the correct answer is:

B. 0.0918