The mean amount purchased by a typical customer at Wal Grocery Store is $23.50,

Question

The mean amount purchased by a typical customer at Wal Grocery Store is $23.50, with a standard deviation of the distribution of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions a. What is likelihood the sample mean at $24.20? b. What is the likelihood the sample mean is greater than $24.20 but less than $25.50? c. Within what limits will 90% of the sample means occur? a) A sample of 36 restaurants revealed that the sample mean number of customers per night was the standard deviation of 9. Construct a 90 percent confidence interval for the population mean of customers per night

Explanation / Answer

Here it is given that mean=23.50 and sd=5

a. P(xbar>=24.20)=P(z>=24.20-23.50/(5/sqrt(50))=P(z>=0.99)=0.5-P(0<=z<=0.99)=0.5-0.3389=0.1611

b. P(24.50<x<25.50)=P(0.99<z<2.83)=0.1588

c. We need to find P(x1<x<x2)=0.90

We know that P(-1.645<z<1.645)=0.90

So z1=x1-mu/sd=x1-23.50/5=-1.645

Hence x1=-1.645*5+23.50=15.275

And x2=1.645*5+23.50=31.725

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