Question

Let 'x' be a random variable that represents the length of time it takes a student to write a term paper for Dr. Adam's Sociology class. After interviewing many students, it was found that 'x' has an approximately normal distribution with a mean of µ = 7.3 hours and standard deviation of ơ = 0.8 hours.

For parts a, b, c, Convert each of the following x intervals to standardized z intervals.

a.) x < 8.1  

z <

b.) x > 9.1  

z >

c.) 5.9 < x < 8.3

< z <

For parts d, e, f, Find the following probabilities: (Use 4 decimal places)

d.) P(x < 8.1) =

e.) P(x > 9.1) =

f.) P(5.9 < x < 8.3) =

Answer 1

Given that, mean (μ) = 7.3 hours and

standard deviation = 0.8 hours

a) x < 8.1

z = (8.1 - 7.3) / 0.8 = 1

=> z < 1

b) x > 9.1

z = (9.1 - 7.3) / 0.8 = 2.25

=> z > 2.25

c) 5.9 < x < 8.3

z = (5.9 - 7.3) / 0.8 = -1.75

z = (8.3 - 7.3) / 0.8 = 1.25

=> -1.75 < z < 1.25

d) P(x < 8.1) = P(z < 1) = 0.8413

=> P(x < 8.1) = 0.8413

e) P(x > 9.1) = P(z > 2.25) = 1 - P(z < 2.25) = 1 - 0.9878 = 0.0122

=> P(x > 9.1) = 0.0122

f) P(5.9 < x < 8.3)

= P(-1.75 < z < 1.25)

= P(z < 1.25) - P(z < -1.75)

= 0.8944 - 0.0401

= 0.8543

=> P(5.9 < x < 8.3) = 0.8543