a) For an audience of 600 people attending a concert, the average time on the journey...

a) For an audience of 600 people attending a concert, the average time on the journey to the concert was 30 minutes, and the standard deviation was 10 minutes. A random sample of 150 audience was taken.

i. What is the probability that the sample mean journey time was less than 32 minutes?

ii. What is the probability that the sample mean exceeds the population mean by more than 1 minute?

Solutions

Expert Solution

Mean, = 30 minutes

Standard deviation, = 10 minutes

Sample size, n = 150

For sampling distribution of mean, P( < A) = P(Z < (A - )/)

= = 30 minutes

= 0.8165

i. P(sample mean journey time was less than 32 minutes) = P( < 32)

= P(Z < (32 - 30)/0.8165)

= P(Z < 2.45)

= 0.9929

ii. P(sample mean exceeds the population mean by more than 1 minute) = P( > 31)

= 1 - P( < 31)

= 1 - P(Z < (31 - 30)/0.8165)

= 1 - P(Z < 1.22)

= 1 - 0.8888

= 0.1112

orchestra answered 2 years ago

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