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?

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.

Step-by-step explanation:

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

In a set with mean and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.9332.

X = 1350

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

has a pvalue of 0.5.

X = 1341

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

has a pvalue of 0.7967

X = 1338

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

has a pvalue of 0.9332.

X = 1365

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.

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i believe it is b and c. (side note, i got the same computer from k-12 lol i’m in 8th grade and this is my first year, will it get

step-by-step explanation: