Question

(1 point) 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 8.88.8 hours of sleep, with a standard deviation of 2.122.12 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 6.56.5 hours of sleep? Round to four decimal places. answer: (b) What is the probability that a visually impaired student gets between 6.96.9 and 10.610.6 hours of sleep? Round to four decimal places. answer: (c) Fourty percent of students get less than how many hours of sleep on a typical day? Round to two decimal places. answer: hours

Answer 1

Answer:

(a) Probability that a visually impaired student gets less than 6.5 hours of sleep is 0.1390.

(b) P(6.9 hours < X < 10.6 hours) = 0.6172

(c) Forty percent of students get less than 8.26 hours of sleep on a typical day.

Step-by-step explanation:

We are given that Researchers found that visually impaired students averaged 8.8 hours of sleep, with a standard deviation of 2.12 hours.

Assume that the number of hours of sleep for these visually impaired students is normally distributed.

Let X = number of hours of sleep for these visually impaired students

So, X ~ Normal($\mu=8.8,\sigma^{2} = 2.12^{2}$)

The z-score probability distribution for normal distribution is given by;

                        Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where,  $\mu$ = mean hours of sleep = 8.8 hours

             $\sigma$ = standard deviation = 2.12 hours

(a) Probability that a visually impaired student gets less than 6.5 hours of sleep is given by = P(X < 6.5 hours)

    P(X < 6.5 hours) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6.5-8.8}{2.12}$ ) = P(Z < -1.085) = 1 - P(Z $\leq$ 1.085)

                                                              = 1 - 0.8610 = 0.1390

The above probability is calculated by looking at the value of x = 1.085 in the z table which will lie in between x = 1.08 and x = 1.09.

(b) Probability that a visually impaired student gets between 6.9 and 10.6 hours of sleep is given by = P(6.9 hours < X < 10.6 hours)

    P(6.9 hours < X < 10.6 hours) = P(X < 10.6 hours) - P(X $\leq$ 6.9 hours)

    P(X < 10.6 hours) = P( $\frac{X-\mu}{\sigma}$ < $\frac{10.6-8.8}{2.12}$ ) = P(Z < 0.85) = 0.80234

    P(X $\leq$ 6.9 hours) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{6.9-8.8}{2.12}$ ) = P(Z $\leq$ -0.896) = 1 - P(Z < 0.896)

                                                               = 1 - 0.81487 = 0.18513

The above probability is calculated by looking at the value of x = 0.85 and x  = 0.896 in the z table which will lie in between x = 0.89 and x = 0.90.

Therefore, P(6.9 hours < X < 10.6 hours) = 0.80234 - 0.18513 = 0.6172

(c) We have to find that Forty percent of students get less than how many hours of sleep on a typical day, that is;

    P(X < $x$) = 0.40    {where $x$ is the hours of sleep}

    P( $\frac{X-\mu}{\sigma}$ < $\frac{x-8.8}{2.12}$ ) = 0.40

         P(Z < $\frac{x-8.8}{2.12}$ ) = 0.40

Now, in the z table the critical value of x for which the probability area is less than 40% is -0.2533, that means;

                  $\frac{x-8.8}{2.12}$  =  -0.2533

                 $x-8.8 = -0.2533 \times 2.12$

                   $x$ = 8.8 - 0.536996 = 8.26

Therefore, Forty percent of students get less than 8.26 hours of sleep on a typical day.