18. In the past, all patrons of a cinema complex have spent an average of $5.00...

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?

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Expert Solution

Solution :

Given that ,

mean = = 5

standard deviation = = 1.80

n = 32

= = 5

= / n = 1.80 / 32 = 0.3182

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

= 1 - P[( - ) / < (4.20 - 5) /0.3182 ]

= 1 - P(z < -2.51 )

= 1 - 0.006 = 0.9940

probability =0.9940.

orchestra answered 1 year ago

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