Answer:
a) 0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.
b) 0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.
c) 0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
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 8.1 minutes and a standard deviation of 2.0 minutes.
This means that
a. between 5 and 10 min
This is the p-value of Z when X = 10 subtracted by the p-value of Z when X = 5.
X = 10
has a p-value of 0.8289
X = 5
has a p-value of 0.0606
0.8289 - 0.0606 = 0.7683
0.7683 = 76.83% probability that a randomly selected emergency call is between 5 and 10 minutes.
b. less than 5 min
p-value of Z when X = 5, which, found from item a, is of 0.0606
0.0606 = 6.06% probability that a randomly received emergency call is of less than 5 min.
c. more than 10 min
1 subtracted by the p-value of Z when X = 10, which, from item a, is of 0.8289
1 - 0.8289 = 0.1711
0.1711 = 17.11% probability that a randomly received emergency call is of more than 10 min.