Question

PART I

Suppose that the mean and standard deviation of the scores on a statistics exam are 78 and 6.11, respectively, and are approximately normally distributed. Calculate the proportion of scores above 74.

	1) 0.8080
	2) 0.1920
	3) 0.2563
	4) We do not have enough information to calculate the value.
	5) 0.7437

PART II

When students use the bus from their dorms, they have an average commute time of 9.969 minutes with standard deviation 3.9103 minutes. Approximately 27.23% of students reported a commute time greater than how many minutes? Assume the distribution is approximately normal.

	1) 12.34
	2) We do not have enough information to calculate the value.
	3) 19.23
	4) 0.7
	5) 7.6

PART III

When students use the bus from their dorms, they have an average commute time of 9.945 minutes with standard deviation 3.2476 minutes. Approximately 42.21% of students reported a commute time less than how many minutes? Assume the distribution is approximately normal.

	1) 7.87
	2) We do not have enough information to calculate the value.
	3) 12.02
	4) 9.31
	5) 10.58

Answer 1

Answer :

Part I:

Suppose that the mean and standard deviation of the scores on a statistics exam are 78 and 6.11, respectively, and are approximately normally distributed. Calculate the proportion of scores above 74.

Given mean, μ = 78

Standard deviation, σ = 6.11

The proportion of score above 74 is

P(X > x) = P[(X - μ)/σ > (74-78)/6.11]

P(X > 74) = P(Z > -0.655)

= 1 - P(Z< 0.655)

= 1 - 0.2546

= 0.7454

As,0.7454 is nearest to the 0.7437

Therefore, proportion of score above 74 is 0.7454

Therefore, the option 5 is correct answer.

Part II :

When students use the bus from their dorms, they have an average commute time of 9.969 minutes with standard deviation 3.9103 minutes. Approximately 27.23% of students reported a commute time greater than how many minutes?

Given,

Average time,μ = 9.969

Standard deviation, σ = 3.9103

P(Z > z) = 27.23%

1 - P(Z < z) = 0.2723   

P( Z < z) = 1- 0.2723

= 0.7277

Z = 0.61 (from cumulative probability of standard normal distribution table)

From Z score formula,

X =Zσ +μ

= -0.61 * 3.9103 + 9.969

= 7.58

~ 7.6

Therefore, approximately 27.23 % of students report a commute time greater than 7.6 minutes.

Therefore, the option 5 is correct answer.

Part III:

Given,

Average time,μ = 9.945

Standard deviation, σ = 3.2476

P(X < x) = 0.4221

P(Z < z) = 0.4221

z = -0.20

X = mean + z* sd = 9.945 -0.20 * 3.2476

= 9.2955

~ 9.30

Therefore, approximately 42.21% of students reported a commit time less than 9.30 minutes.

Therefore, the option 4 is correct answer.