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The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a standard deviation of 3 minutes. Based upon the above information and numerically justified, would you be surprised if it took less than one minute to find a parking space

User Brenzy
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Answer:

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by


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

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.

Also, a probability is unusual if it is lesser than 5%. If it is unusual, it is surprising.

In this problem:

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 7 minutes and a standard deviation of 3 minutes, so
\mu = 7, \sigma = 3.

We need to find the probability that it takes less than one minute to find a parking space.

So we need to find the pvalue of Z when
X = 1


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


Z = (1 - 7)/(3)


Z = -2


Z = -2 has a pvalue of 0.0228.

There is a 2.28% probability that it takes less than one minute to find a parking space. Since this probability is smaller than 5%, you would be surprised to find a parking space so fast.

User Dominic Clifton
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