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 8.85 hours of sleep, with a standard deviation of 2.19 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 6.5 hours of sleep?

answer:

(b) What is the probability that a visually impaired student gets between 6.8 and 10.37 hours of sleep?

answer:

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

answer: hours

Answer #1

Solution :

Given that ,

mean = = 8.85

standard deviation = = 2.19

**a)** P(x < 6.5) = P[(x -
) /
< (6.5 - 8.85) /2.19 ]

= P(z < -1.07)

= 0.1423

**answer = 0.1423**

**b)** P(6.8 < x < 10.37) = P[(6.8 -
8.85)/2.19 ) < (x -
) / <
(10.37 - 8.85) /2.19 ) ]

= P(-0.94 < z < 0.69)

= P(z < 0.69) - P(z < -0.94)

= 0.7549 - 0.1736

0.5813

**answer = 0.5813**

**c)** Using standard normal table ,

P(Z < z) = 20%

P(Z < z) = 0.2

P(Z < -0.84) = 0.2

z = -0.84

Using z-score formula,

x = z * +

x = -0.84 * 2.19 + 8.85 = 7.01

answer = 7.01 hours

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