In the past, patrons of a cinema complex have spent an average of $5.00 for popc

Question

In the past, 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 the amounts spent on popcorn AND other snacks are normally distributed,

What is the probability that a patron chosen at random spends more than $4.20?

What is the mean AND standard deviation of the sampling distribution of the sample means of samples of 25 patrons?

If a random sample of 25 patrons is taken, what is the probability that the average expenditure of this sample is greater than $4.20?

Explanation / Answer

Ans:

Given that

mean=5

standard deviation=1.8

1)

z=(4.2-5)/1.8=-0.44

P(x>4.2)=P(z>-0.44)=P(z<0.44)=0.6700

2)sampling distribution of smaple means:

mean=5

standard deviatiom=1.8/sqrt(25)=1.8/5=0.36

3)

z=(4.2-5)/0.36=-0.8/0.36=-2.22

P(z>-2.22)=P(z<2.22)=0.9861

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