Question

A journal published a study of the lifestyles of visually impaired students. Using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. These vis impaired students had a mean of 9.3 hours and a standard deviation of 1.88 hours. Assume that the distribution of the number of hours of sleep for this group of students is approximately normal. Below what hours 85% of the visually impaired students have their sleep? [Answer to 2 decimal places] A: 8.63 B: 9.88 C: 11.25 OD: 13.12 OE: 14.84 OF: 14.96 Submit Answer Tries 0/4 What is the chance that a randomly selected visually impaired student has at least 8.5 hours of sleep? [Answer to 4 decimal places] A: 0.1751 B: 0.2158 C: 0.5779 OD: 0.6393|E: 0.6648|OF: 0.9167| Submit Answer Tries 0/4 What is the inter-quartile range of sleep hours among visually impaired students. [Answer to 2 decimal places] A: 2.39 B: 2.51 OC: 2.54 OD: 2.87 OE: 2.89 OF: 3.92

Answer 1

Part a)

X ~ N ( µ = 9.3 , σ = 1.88 )
P ( X < x ) = 85% = 0.85
To find the value of x
Looking for the probability 0.85 in standard normal table to calculate Z score = 1.0364
Z = ( X - µ ) / σ
1.0364 = ( X - 9.3 ) / 1.88
X = 11.2484 ≈ 11.25
P ( X < 11.25 ) = 0.85

C: 11.25

part b)

X ~ N ( µ = 9.3 , σ = 1.88 )
P ( X > 8.5 ) = 1 - P ( X < 8.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 8.5 - 9.3 ) / 1.88
Z = -0.4255
P ( ( X - µ ) / σ ) > ( 8.5 - 9.3 ) / 1.88 )
P ( Z > -0.4255 )
P ( X > 8.5 ) = 1 - P ( Z < -0.4255 )
P ( X > 8.5 ) = 1 - 0.3352
P ( X > 8.5 ) = 0.6648

E: 0.6648

Part c)

X ~ N ( µ = 9.3 , σ = 1.88 )
P ( a < X < b ) = 0.5
Dividing the area 0.5 in two parts we get 0.5/2 = 0.25
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.25
Area above the mean is b = 0.5 + 0.25
Looking for the probability 0.25 in standard normal table to calculate Z score = -0.6745
Looking for the probability 0.75 in standard normal table to calculate Z score = 0.6745
Z = ( X - µ ) / σ
-0.6745 = ( X - 9.3 ) / 1.88
a = 8.0319
0.6745 = ( X - 9.3 ) / 1.88
b = 10.5681
P ( 8.0319 < X < 10.5681 ) = 0.5

Q1 = 8.03

Q2 = 10.57

IQR = Q3 - Q1 = 2.54