Question

In: Statistics and Probability

how do you get the answers using a calculator. the answers is 0.9940. my professor gave...

how do you get the answers using a calculator. the answers is 0.9940. my professor gave us  a review with answers but didn't explain the answers. she wrote 2nd VARS (2) normal cdf ( lower, upper, u, sd)

18. In the past, all patrons of a cinema complex have spent an average of $5.00 for popcorn and other snacks, with a standard deviation of $1.80. If a random sample of 32 patrons is taken, what is the probability that the mean expenditure of this sample is greater than $4.20?

Solutions

Expert Solution

Solution :

Given that ,

mean = = 5.00

standard deviation = = 1.80

n = 32

= = 5.00 and

= / n = 1.80 / 32 = 0.3182

P( > 4.20) = 1 - P( < 4.20)

= 1 - P(( - ) / < (4.20 - 5.00) / 0.3182)

= 1 - P(z < -2.51)

= 1 - 0.0060

= 0.9940

Probability = 0.9940


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