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 8.818.81 hours of sleep, with a standard deviation of 1.61.6 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 6.76.7 hours of sleep?

(b)  What is the probability that a visually impaired student gets between 6.66.6 and 7.847.84 hours of sleep?

(c)  Thirty percent of students get less than how many hours of sleep on a typical day?

Answer 1

Solution:- Given that mean = 8.1, sd = 1.6
(a) P(X < 6.7) = P((X-mean)/sd < (6.7-8.1)/1.6)
= P(Z < -0.875)
= 1 − P(Z < 0.875)
= 1 − 0.8106
= 0.1894

Now we can find P ( Z<0.875 ) by using the standard normal table.

Z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5	0.504	0.508	0.512	0.516	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.591	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.648	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.67	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.695	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.719	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.758	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.791	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078	0.8106	0.8133

(b) P(6.6 < X < 7.84) = P((6.6-8.1)/1.6 < Z < (7.84-8.1)/1.6)
= P(-0.9375 < Z < -0.1625)
= P(Z < −0.1625) − P(Z < −0.9375)
= 0.4364 - 0.1736
= 0.2628

(c) P(X < Z) = 0.30, Z = -0.5244

X = mean + Z*Sd = 8.1 - (0.5244*1.6) = 7.26