Question

Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.6 hours of sleep, with a standard deviation of 1.21 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

(a) What is the probability that a visually impaired student gets less than 6.9 hours of sleep? answer: 0.0129

(b) What is the probability that a visually impaired student gets between 6.6 and 10.53 hours of sleep? answer: .7728

(c) Fourty percent of students get less than how many hours of sleep on a typical day? answer:

Answer 1

Solution :

Given that ,

mean = = 9.6

standard deviation = =1.21

P(X<6.9 ) = P[(X- ) / < (6.9-9.6) / 1.21]

= P(z < -2.23)

Using z table

= 0.0129

probability=0.0129

(b)

P(6.6< x < 10.53) = P[(6.6-9.6) / 1.21< (x - ) / < (10.53-9.6) / 1.2 )]

= P( -2.48< Z < 0.775)

= P(Z < 0.775) - P(Z < - 2.48 )

Using z table   

= 0.7808-0.0066

probability= 0.7742

(d)

Using standard normal table,

P(Z < z) =40 %

= P(Z < z) = 0.40  

= P(Z < -0.25 ) = 0.40

z = - 0.25 Using standard normal table,

Using z-score formula  

x= z * +

x= - 0.25*1.21+9.6

x= 9.2975

x=9