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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(less than 6.2 hours of sleep)

$P( x < 6.2) = P( z < \displaystyle\frac{6.2 - 9.54}{2.62}) = P(z < -1.2748)$

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.

Answer 2

Answer:

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

Step-by-step 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 = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ < $\frac{6.2-9.54}{2.62}$ ) = 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.