The mean useful life of car batteries is 48 months. They have a standard deviation of 3. Assume the useful life of batteries is normally distributed. a. Calculate the percent of batteries with a useful life of less than 45 months. (Round your answer to the nearest tenth percent.) b. Calculate the percent of batteries that will last longer than 54 months. (Round your answer to the nearest tenth percent.)

a) 15.9%

b) 2.3%

Step-by-step explanation:

We solving using z score formula. This formula is given as:

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a. Calculate the percent of batteries with a useful life of less than 45 months. (Round your answer to the nearest tenth percent.)

x = 45, μ = 48, σ = 3

z = 45 - 48/3

= -1

Probability value from Z-Table:

The probability of batteries with a useful life of less than 45 months so

P(x<45) = 0.15866

Converting to percentage

0.15866 × 100 = 15.866

Approximately to the nearest tenth of a percent = 15.9%

b. Calculate the percent of batteries that will last longer than 54 months. (Round your answer to the nearest tenth percent.)

x = 58, μ = 48, σ = 3

z = 58 - 48/3

= 2

Probability value from Z-Table:

P(x<54) = 0.97725

P(x>54) = 1 - P(x<54)

= 0.02275

The probability of batteries that will last longer than 54 months is 0.02275

Converting to percentage

0.02275 × 100

= 2.275%

Approximately to the nearest tenth of a percentage = 2.3%

answer: y=0 if that is what u r looking for

the answer is 40.

proof: