Yes 10 pts) Using diaries for many weeks, a study on the lifestyles of visually

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Yes 10 pts) Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many festyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.83 hours of sleep, with a standard deviation of 1.93 hours. Assume that the number of hours of sleep tor th normally distributed. (a) What is the probability that a visually impaired student gets at most 6 hours of sleep? Express your answer as a percent rounded to 2 decimal places e g 1 2396 Do not include the % symbol in your answer. b) What is the probability that a visually impaired student gets between 6.6 and 9.83 hours of sleep? Express your answer as a percent rounded to 2 decimal places, e g 1 23% Do not include the % symbol in your answer (e) What is the probability that a visually impaired student gets at least 8.6 hours of sleep? Express your answer as a percent rounded to 2 decimal places e g 1.23% Do not include the % symbol in your answer. (d) what is the sleep time that cuts off the top 25% of sleep hours? Round your answer to 2 decinal places Answer hours (e) If 150 visually impaired students were studied, how many students would you expect to have sleep times of more than 9.83 hours? Round to the nearest whole number t) A school district wants to give additional assistance to visualily impaired students with sleep times at the first quartile and lower. What would be the maximum sleep time to be recommended tor additional assistance? Round your answer to 2 decumal places Answer hours O Type here to search 8/12/2018

Explanation / Answer

Ans:

Given that

mean=9.83

standard deviation=1.93

a)

z=(6-9.83)/1.93=-1.984

P(z<=-1.984=0.0236

2.36%

b)

z(6.6)=(6.6-9.83)/1.93=-1.673

z(9.83)=0

P(-1.67<z<0)=P(z<0)-P(z<-1.67)=0.5-0.0471=0.4529

45.29%

c)

z=(8.6-9.83)/1.93=-0.637

P(z>=-0.637)=0.7380

73.80%

d)P(Z>=z)=0.25

z=0.6745

cut off=9.83+0.6745*1.93=11.13

e)z=0

P(z>0)=0.5

Expected number of student who sleep more than 983 hrs=0.5*150=75

f)First quartile means

P(Z<=z)=0.25

z=-0.6745

maximum sleep time=9.83-0.6745*1.93=8.53 hrs

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