Question

The lives of certain extra-life light bulbs are normally distributed with a mean equal to 1350 hours and a standard deviation equal to 18 hours1. What percentage of bulbs will have a life between 1350 and 1377 hr?2. What percentage of bulbs will have a life between 1341 and 1350 hr?3. What percentage of bulbs will have a life between 1338 and 1365 hr?4. What percentage of bulbs will have a life between 1365 and 1377 hr?

Answer 1

1. 43.32% of bulbs will have a life between 1350 and 1377 hr.

2. 19.15% of bulbs will have a life between 1341 and 1350 hr.

3. 54.53% of bulbs will have a life between 1338 and 1365 hr.

4. 13.65% of bulbs will have a life between 1365 and 1377 hr.

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 = 1350, \sigma = 18`

1. What percentage of bulbs will have a life between 1350 and 1377 hr?

This is the pvalue of Z when X = 1377 subtracted by the pvalue of Z when X = 1350.

X = 1377

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

`Z = (1377 - 1350)/(18)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332.

X = 1350

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

`Z = (1350 - 1350)/(18)`

`Z = 0`

`Z = 0` has a pvalue of 0.5.

So 0.9332 - 0.5 = 0.4332 = 43.32% of bulbs will have a life between 1350 and 1377 hr.

2. What percentage of bulbs will have a life between 1341 and 1350 hr?

This is the pvalue of Z when X = 1350 subtracted by the pvalue of Z when X = 1341.

X = 1350

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

`Z = (1350 - 1350)/(18)`

`Z = 0`

`Z = 0` has a pvalue of 0.5.

X = 1341

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

`Z = (1341 - 1350)/(18)`

`Z = -0.5`

`Z = -0.5` has a pvalue of 0.3085.

So 0.5 - 0.3085 = 0.1915 = 19.15% of bulbs will have a life between 1341 and 1350 hr.

3. What percentage of bulbs will have a life between 1338 and 1365 hr?

This is the pvalue of Z when X = 1365 subtracted by the pvalue of Z when X = 1338.

X = 1365

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

`Z = (1365 - 1350)/(18)`

`Z = 0.83`

`Z = 0.83` has a pvalue of 0.7967

X = 1338

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

`Z = (1338 - 1350)/(18)`

`Z = -0.67`

`Z = -0.67` has a pvalue of 0.2514.

So 0.7967 - 0.2514 = 0.5453 = 54.53% of bulbs will have a life between 1338 and 1365 hr.

4. What percentage of bulbs will have a life between 1365 and 1377 hr?

This is the pvalue of Z when X = 1377 subtracted by the pvalue of Z when X = 1365.

X = 1377

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

`Z = (1377 - 1350)/(18)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332.

X = 1365

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

`Z = (1365 - 1350)/(18)`

`Z = 0.83`

`Z = 0.83` has a pvalue of 0.7967.

So 0.9332 - 0.7967 = 0.1365 = 13.65% of bulbs will have a life between 1365 and 1377 hr.