Question

Ajoumal beshed a study of the eyes of visually impaired students Ungdanes, the dents kastrack of several are doing ber of hours of p laned in a local day. These visually impaired students had a man of 18 hours and a standard deviation of 213 hours Assume that the trution of the number of hours of sleep for the group of students is approximately omal Complete parts a through a. Find P 12) Pe 12)- (Round to four decimal places as needed) Find P8** 101 PBX10-Round to our complet ed c. Find the value a for which Pixa)02 Round to two decimal places as needed)

Answer 1

a.

μ = 9.18, σ = 2.13

We need to compute Pr(X≥12).

The corresponding z-value needed to be computed is:

$Z = \frac{X - \mu}{\sigma} = \frac{ 12 - 9.18}{ 2.13} = 1.3239$

Therefore, we get that

$\Pr(X \geq 12) = \Pr\left(Z \ge \frac{ 12 - 9.18}{ 2.13}\right) = \Pr(Z \ge 1.3239)\\ \\= 1 - 0.9072 = 0.0928$

b.

μ = 9.18, σ = 2.13

We need to compute Pr( 8 ≤ X ≤ 10).

The corresponding z-values needed to be computed are:

X1 -μ Ζ = 8 – 9.18 - = -0.554 2.13

Z, X2- 10 – 9.18 -= 0.385 2.13

Therefore, we get:

$\Pr(8 \leq X \leq 10) = \Pr\left(\frac{ 8 - 9.18}{ 2.13} \le Z \le \frac{ 10 - 9.18}{ 2.13}\right)\\ \\= \Pr(-0.554 \le Z \le 0.385)= \Pr(Z \le 0.385) - \Pr(Z \le -0.554) \\ \\= 0.6499 - 0.2898 = 0.3601$

c.

We need to find a such that P(X<a) = 0.2

> qnorm(0.2) [1] -0. 8416212

$Z = \frac{a - \mu}{\sigma} \Rightarrow -0.8416212 = \frac{a - 9.18}{2.13}\\ \\ a = -0.8416212 * 2.13 + 9.18 = 7.387347 \approx 7.39$