Question

Consider the variable x = time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of x is well approximated by a normal curve with mean 50 minutes and standard deviation 5 minutes. (Use a table or technology.)

(a)

If 55 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? (Round your answer to four decimal places.)

(b)

How much time (in minutes) should be allowed for the exam if you wanted 90% of the students taking the test to be able to finish in the allotted time? (Round your answer to one decimal place.)

min

(c)

How much time (in minutes) is required for the fastest 25% of all students to complete the exam? (Round your answer to one decimal place.)

min

Answer 1

Solution:

Given:  x = time required for a college student to complete a standardized examination.

the distribution of x is well approximated by a normal curve with mean 50 minutes and standard deviation 5 minutes.

Part a) If 55 minutes is allowed for the examination, what proportion of students at this university would be unable to finish in the allotted time?

That is find:
P( X > 55) = ...........?

Find z score for x = 55

Thus we get:

P( X > 55) = P( Z > 1.00 )

P( X > 55) = 1 - P( Z < 1.00 )

Look in z table for z = 1.0 and 0.00 and find corresponding area.

P( Z< 1.00) = 0.8413

Thus

P( X > 55) = 1 - P( Z < 1.00 )

P( X > 55) = 1 - 0.8413

P( X > 55) = 0.1587

Part b) How much time (in minutes) should be allowed for the examination if you wanted 90% of the students taking the test to be able to finish in the allotted time?

That is find x value such that:

P( X < x ) =90%

P( X < x ) =0.90

Thus find z value such that:
P(Z < z) = 0.90

Look in z table for area = 0.9000 or its closest area and find z value:

Area 0.8997 is closest to 0.9000 and it corresponds to 1.2 and 0.08

thus z = 1.28

Now use following formula to find x value:

minutes

Part c) How much time (in minutes) is required for the fastest 25% of all students to complete the examination?

Time (in minutes) is required for the fastest 25% of all students means , students who completed examination before the time or took less time than others.

thus find x value such that:

P( X < x ) =25%

P( X < x ) =0.25

Thus find z value such that:
P(Z < z) = 0.25

Look in z table for area = 0.2500 or its closest area and find z value:

Area 0.2514 is closest to 0.2500 and it corresponds to -0.6 and 0.07

Thus z = -0.67

Now use following formula to find x value:

minutes.