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 were obtained on a typical day. Researchers found that visually impaired students averaged 9.87 hours of sleep, with a standard deviation of 1.1 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 6.1 hours of sleep?
b. What is the probability that a visually impaired student gets between 6.5 and 7.89 hours of sleep?

Answer 1

a) 0.0003

b) 0.035

Explanation:

We are given the following information in the question:

Mean, μ = 9.87 hours

Standard Deviation, σ = 1.1 hours

We are given that the distribution of hours of sleep is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(student gets less than 6.1 hours of sleep)

P(x < 6.1)

`P( x < 6.1) = P( z < \displaystyle(6.1 - 9.87)/(1.1)) = P(z < -3.4272)`

Calculation the value from standard normal z table, we have,

`P(x < 6.1) = 0.0003`

c)P(student gets between 6.5 and 7.89 hours of sleep)

`P(6.5 \leq x \leq 7.89) = P(\displaystyle(6.5 - 9.87)/(1.1) \leq z \leq \displaystyle(7.89-9.87)/(1.1)) = P(-3.063 \leq z \leq -1.8)\\\\= P(z \leq -1.8) - P(z < -3.063)\\= 0.036 - 0.001 = 0.035 = 3.5\%`

`P(6.1 \leq x \leq 7.89) = 3.5\%`