(1 point) Using diaries for many weeks, a study on the lifestyles of visually im
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(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 9.25 hours of sleep, with a standard deviation of 2.15 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 at most 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 Answer: (b) What is the probability that a visually impaired student gets between 6.5 and 9.84 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. Answer: (c) What is the probability that a visually impaired student gets at least 8 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. Answer (d) What is the sleep time that cuts off the top 1 1 .2% of sleep hours? Round your answer to 2 decimal places Answer: hours (e) If 700 visually impaired students were studied, how many students would you expect to have sleep times of more than 9.84 hours? Round to the nearest whole number Answer students (f) A school district wants to give additional assistance to visually impaired students with sleep times at the first quartile and lower. What would be the maximum sleep time to be recommended for additional assistance? Round your answer to 2 decimal places Answer: hoursExplanation / Answer
a) as z score =(X-mean)/std deviaiton
therefore probability=P(X<6) =P(X<6)=P(Z<(6-9.25)/2.15)=P(Z<-1.5116)=0.0653~ 6.53%
b) P(6.5<X<9.84)=P(-1.2791<Z<0.2744)=0.6081-0.1004 =0.5077 ~ 50.77%
c)P(X>8)=1-P(X<8)=1-P(Z<-0.5814)=1-0.2805 =0.7195 ~ 71.95%
d)for 11.2% top most ; at 88.8 percentile ; z score =1.2160
threfore corresponding sleep time =mean +z*Std deviaiton =9.25+1.2160*2.15 =11.86 Hours
e)for probability of sleeping more then 9.84 hours =P(X>9.84)=1-P(X<9.84)=1-P(Z<0.2744)=1-0.6081=0.3919
hence number of students =np=700*0.3919 =~274
f)for first quartile ; at 25h percentile z =-0.6745
threfore corresponding sleep time =mean +z*Std deviaiton =9.25-0.6745*2.15 =7.80 Hours
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