In: Statistics and Probability
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?
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.