Question

Question 1

An office will increase salary to its top 8% employees on the basis of a performance score the office created for each employee. The performance score is approximately normal with mean 82.5 and standard deviation 9.25.

How high must an employee score in order to qualify for increase in the salary?

Group of answer choices

87.50

85.75

95.50

82.50

Question 2

The performance score of employees in an office is approximately normal with mean 82.5 and standard deviation 9.25.

What is the third quartile (Q3) of the performance score of the employees?

(Hints: Recall that the third quartile is a score with 75% data values below it and 25% data values above it)

Group of answer choices

9.25

88.74

82.50

76.26

Question 3

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age with a mean of 93 seconds and a standard deviation of 16 seconds. How fast does a man have to run to be in the top 5% of runners (quickest runner)?

Group of answer choices

119.32 seconds

66.68 seconds

109.32 seconds

93.68 seconds

Question 4

The typical college freshman spends an average (mean) of 150 minutes per day with a standard deviation of 50 minutes on social media. The distribution of time on social media is known to be Normal.

What is the probability that students will spend between 140 minutes and 200 minutes on social media?

Group of answer choices

0.22

0.82

0.62

0.42

Question 5

A company produces packets of soap powder labeled “Giant Size 32 Ounces.” The actual weight of soap powder in such a box has a Normal distribution with a mean of 33 oz and a standard deviation of 0.7 oz. The company says a box of soap is considered underweight if it weighs less than 32 oz.

What is the probability that a box of soap is underweight (i.e., weigh less than 32 oz)?

Group of answer choices

0.7066

0.0766

0.7

0.5066

Answer 1

Solution :

Given that,  

Question 1

mean = = 82.5

standard deviation = = 9.25

Using standard normal table ,

P(Z > z) = 8%

1 - P(Z < z) = 0.08

P(Z < z) = 1 - 0.08

P(Z < 1.405) = 0.92

z = 1.405

Using z-score formula,

x = z * +

x = 1.405 * 9.25 + 82.5 = 95.50

95.50 high must an employee score in order to qualify for increase in the salary

Question 2

mean = = 85.5

standard deviation = = 9.25

Using standard normal table ,

P(Z < z) = 75%

P(Z < 0.6745) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.6745 * 9.25 + 82.5 = 88.74

88.74 the third quartile (Q3) of the performance score of the employees

Question 3

mean = = 93

standard deviation = = 16

Using standard normal table ,

P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 16 + 93 = 119.32

119.32 seconds

Question 4

mean = = 150

standard deviation = = 50

P(140 < x < 200) = P[(140 - 150)/ 50) < (x - ) /  < (200 - 150) / 50) ]

= P(-0.2 < z < 1)

= P(z < 1) - P(z < -0.2)

= 0.8413 - 0.4207

= 0.42

Probability = 0.42

Question 5

mean = = 33

standard deviation = = 0.7

P(x < 32) = P[(x - ) / < (32 - 33) / 0.7]

= P(z < -1.4286)

= 0.0766

Probability = 0.0766