Question

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between −2.23 and −1.69 and draw a sketch of the region.

Answer 1

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

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.

Mean of 0°C and a standard deviation of 1.00°C.

This means that
`\mu = 0, \sigma = 1`

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

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

`Z = (-1.69 - 0)/(1)`

`Z = -1.69`

`Z = -1.69` has a p-value of 0.0455

X = -2.23

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

`Z = (-2.23 - 0)/(1)`

`Z = -2.23`

`Z = -2.23` has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch: