The average time a person spends in each visit to an online social networking se

Question

The average time a person spends in each visit to an online social networking service is 58 minutes. The standard deviation is 15 minutes. If a visitor is selected at random, find the probability that they will spend the time shown on the networking service. Assume the times are normally distributed.

Refer to the table of values as needed. If necessary, round intermediate calculations to the nearest hundredth.

At least 176 minutes

The probability that a randomly selected visitor spends at least 176 minutes per visit is________%

Can you show me how you got your answer?

Explanation / Answer

First Part )

z = (58-58)/15 = 0

Probability that they will spend the time shown on the networking service = 0.5 Answer

Second Part)

Z(X>=176) = ( 176 -58 ) / 15= 7.8

P-value = 1 - 0.99999 ~ 0.00001 = 0.001% Answer

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