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 were obtained on a typical day. Researchers found that visually impaired students averaged 9.87 hours of sleep, with a standard deviation of 1.1 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 6.1 hours of sleep? What is the probability that a visually impaired student gets between 6.5 and 7.89 hours of sleep? b.

Answer 1

Let the random variable X denotes the number of hours of sleep and we are given that X~N(mean,variance) ;

where mean = 9.87 and variance= (1.1)²

^{a) Probability that a visually impaired student gets less than 6.1 hour of sleep is given by}

^{P (X<6.1)= P(((X-9.87)/1.1) < (6.1-9.87)/1.1)}

^{=P( Z < - 3.42727) = P(Z > 3.42727) { Since Z is symmetric distribution}}

^{= 0.5 - P( 0 < Z < 3.42727 ) = 0.5 - 0.4997 = 0.0003 : Ans}

^{b) Acc to ques we need to find : P( 6.5 < X < 7.89 )}

^{= P( (6.5 - 9.87)/1.1 < Z < ( 7.89 - 9.87 )/1.1 )}

^{= P ( - 3.06 < Z < - 1.8 ) = P(Z < -1.8) - P( Z < -.3.06) ..........(1)}

Now ,Consider P( Z< -1.8) = P( Z > 1.8 ) = 0.5 - P ( 0 < Z < 1.8)

= 0.5 - 0.3599 = 0.1401

For P( Z < -3.06) = P( Z> 3.06) = 0.5- P (0 < Z < 3.06)

= 0.5 - 0.4989 = 1.1 * 10^-3  = 0.0011

Now putting these values in (1), we get

^{P( - 3.06 < Z < - 1.8 ) = 0.1401 - 0.0011 = 0.139 : Ans}