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 9.54 hours of sleep, with a standard deviation of 2.62 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 6.2 hours of sleep

Answer 1

(a) Probability that a visually impaired student gets less than 6.2 hours of sleep is 0.10204 .

Explanation:

We are given that Researchers found that visually impaired students averaged 9.54 hours of sleep, with a standard deviation of 2.62 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

Let X = number of hours of sleep

So, X ~ N(
`\mu = 9.54,\sigma^(2)=2.62^(2)`)

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean = 9.54 hours

`\sigma` = standard deviation = 2.62 hours

So, the probability that a visually impaired student gets less than 6.2 hours of sleep is given by = P(X < 6.2)

P(X < 6.2) = P(
`(X-\mu)/(\sigma)` <
`(6.2-9.54)/(2.62)` ) = P(Z < -1.27) = 1 - P(Z
`\leq` 1.27)

= 1 - 0.89796 = 0.10204

Therefore, the probability that a visually impaired student gets less than 6.2 hours of sleep is 0.10204.

Answer 2

0.1012 is the probability that a visually impaired student gets less than 6.2 hours of sleep.

Explanation:

We are given the following information in the question:

Mean, μ = 9.54 hours

Standard Deviation, σ = 2.62 hours

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

Formula:

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

P(less than 6.2 hours of sleep)

`P( x < 6.2) = P( z < \displaystyle(6.2 - 9.54)/(2.62)) = P(z < -1.2748)`

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

`P(x < 6.2) =0.1012= 10.12\%`

0.1012 is the probability that a visually impaired student gets less than 6.2 hours of sleep.