The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 68 customers, answer the following questions a. What is the likelihood the sample mean is at least $30.00?
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
c. Within what limits will 90 percent of the sample means occur?

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

a. What is the likelihood the sample mean is at least $30.00?

This is 1 subtracted by the pvalue of Z when X = 30. So

By the Central Limit Theorem, we have that:

has a pvalue of 0.9984

1 - 0.9984 = 0.0016

0.0016 = 0.16% probability that the sample mean is at least $30.00.

b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?

This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So

From a, when X = 30, Z has a pvalue of 0.9984

When X = 26.5

has a pvalue of 0.1190

0.9984 - 0.1190 = 0.8794

0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.

c. Within what limits will 90 percent of the sample means occur?

Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645

Lower bound:

Upper Bound:

90% of sample means will occur between $26.1 and $28.9.

answer: the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

step-by-step explanation:

the given rate of interest i= 85%

in calculating the monthly payment for a loan with an interest rate compounded monthly, the value of the interest rate per period we use in the formula must be in decimal form.

to convert it in decimal, we divide it by 100, we get

hence, the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

$450 is the answer.

step-by-step explanation:

30% of 450 = 135

450-135=315