Question

The lifetime of a certain type of battery is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the probability that a randomly selected battery will last between 950 and 1000 (round answers to three decimal places, example 0.xxx)?

The lifetime of a certain type of battery is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the probability that a randomly selected battery will last 1070 hours or less (round answers to three decimal places, example 0.xxx)?

The lifetime of a certain type of battery is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the probability that a randomly selected battery will last 930 or less (round answers to three decimal places, example 0.xxx)?

The lifetime of a certain type of battery is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the probability that a randomly selected battery will last between 1000 and 1150 hours (round answers to three decimal places, example 0.xxx)?

Answer 1

This is a normal distribution question with

a) P(950.0 < x < 1050.0)=?

This implies that
P(950.0 < x < 1050.0) = P(-0.5 < z < 0.5) = P(Z < 0.5) - P(Z < -0.5)
P(950.0 < x < 1050.0) = 0.6914 - 0.3085
P(950.0 < x < 1050.0) = 0.3829
b) P(x < 1070.0)=?
The z-score at x = 1070.0 is,

z = 0.7
This implies that
P(x < 1070.0) = P(z < 0.7) = 0.75804
c) P(x < 930.0)=?
The z-score at x = 930.0 is,

z = -0.7
This implies that
P(x < 930.0) = P(z < -0.7) = 0.242
d) P(1000.0 < x < 1150.0)=?

This implies that
P(1000.0 < x < 1150.0) = P(0.0 < z < 1.5) = P(Z < 1.5) - P(Z < 0.0)
P(1000.0 < x < 1150.0) = 0.9332 - 0.5
P(1000.0 < x < 1150.0) = 0.4332
PS: you have to refer z score table to find the final probabilities.
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