Question

(3 marks) A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year. The battery lives are assumed normally distributed. (a) What is the probability that a given battery will last between 2.3 and 3.6 years

Answer 1

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year

This means that
`\mu = 3, \sigma = 0.5`

What is the probability that a given battery will last between 2.3 and 3.6 years?

This is the p-value of Z when X = 3.6 subtracted by the p-value of Z when X = 2.3. So

X = 3.6

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

`Z = (3.6 - 3)/(0.5)`

`Z = 1.2`

`Z = 1.2` has a p-value of 0.8849

X = 2.3

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

`Z = (2.3 - 3)/(0.5)`

`Z = -1.4`

`Z = -1.4` has a p-value of 0.0808

0.8849 - 0.0808 = 0.8041

0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years