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.

19.side- angle- side

20. angle- side- angle

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}