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Sagot :
Answer:
0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.
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 7.2 minutes and a standard deviation of 2.1 minutes.
This means that [tex]\mu = 7.2, \sigma = 2.1[/tex]
For a randomly received emergency call, find the probability that the response time is between 3 and 9 minutes.
This is the pvalue of Z when X = 9 subtracted by the pvalue of Z when X = 3.
X = 9
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9 - 7.2}{2.1}[/tex]
[tex]Z = 0.86[/tex]
[tex]Z = 0.86[/tex] has a pvalue of 0.8051
X = 3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 7.2}{2.1}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
0.8051 - 0.0228 = 0.7823
0.7823 = 78.23% probability that the response time is between 3 and 9 minutes.
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