The amount of money collected by a snack bar at a large university has been recorded daily for the past five years. Records indicate that the mean daily amount collected is $ 2 comma 350 and the standard deviation is $ 400. The distribution is skewed to the right due to several high volume days (including football game days). Suppose that 100 days were randomly selected from the five years and the average amount collected from those days was recorded. Which of the following describes the sampling distribution of the sample mean?
A) normally distributed with a mean of $1950 and a standard deviation of $500
B) normally distributed with a mean of $1950 and a standard deviation of $50
C) normally distributed with a mean of $195 and a standard deviation of $50
D) skewed to the right with a mean of $1950 and a standard deviation of $500

Note: In this question, according to the data given in the question, options are wrong and should be neglected.

Correct Answer must be: Normally distributed with a mean of $ 2,350 and a standard deviation of $40    

Step-by-step explanation:

Here in this question, the mean = 2,350

with standard deviation of = 400

Furthermore, in this question the sample size = 100 = n

and n > 30 = greater than 30, it means it is large enough.

So the sampling distribution of sample mean is normally distributed with mean = 2350 and

Standard Deviation = SD given/

Standard deviation = 400/

Standard Deviation = 40

Hence, the answer is not included in the options given.

Note: In this question, according to the data given in the question, options are wrong and should be neglected.

Correct Answer must be: Normally distributed with a mean of $ 2,350 and a standard deviation of $40

you already gave the answer

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;