Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 0, \sigma = 1[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{-1.69 - 0}{1}[/tex]
[tex]Z = -1.69[/tex]
[tex]Z = -1.69[/tex] has a p-value of 0.0455
X = -2.23
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{-2.23 - 0}{1}[/tex]
[tex]Z = -2.23[/tex]
[tex]Z = -2.23[/tex] 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:
