# QNT 275 Final Exam 2017 – Probability Distribution Problems

For the probability distribution of a discrete random variable x, the sum of the probabilities of all values of x must be:

Equal 1

Explanation: A sum of all values in a probability distribution will always add up to 1, period.

The following table lists the probability distribution of a discrete random variable x:

X:  2, 3, 4, 5, 6, 7, 8

P(x): 0.15, 0.29, 0.26, 0.13, 0.09, 0.06, 0.02

The mean of the random variable x is:

First, multiply each value of x by its probability P(x), then add all the values together: (2*0.15) + (3*0.29) + (4*0.26) + (5*0.13) + (6*0.09) + (7*0.06) + (8*0.02)

3.98

The standard deviation of the random variable x, rounded to three decimal places, is:

1.497

The daily sales at a convenience store produce a distribution that is approximately normal with a mean of 1350 and a standard deviation of 144.

In your intermediate calculations, round z-values to two decimal places.

The probability that the sales on a given day at this store are more than 1405, rounded to four decimal places, is:

Find the z score (observation – mean)/ standard deviation or (1405 – 1350) / 144

0.38194

The probability that the sales on a given day at this store are less than 1305, rounded to four decimal places, is:

(1305 – 1350) / 144

0.3125

The width of a confidence interval depends on the size of the:

margin of error

Explanation: The confidence interval is the bounds of the margin of error in a statistical test.

A sample of size 82 from a population having standard deviation = 52 produced a mean of 255.00. The 95% confidence interval for the population mean:

Use the following interval Confidence interval = m +/- (t(α, N-1)*SEM)

Lower limit: 243.57

Upper limit: 266.43

• Student: Lonny McDowell
• Textbook: QNT 275 Statistics Fundamentals
• Course: QNT 275 Final Exam 2017