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 8.82 hours of sleep, with a standard deviation of 2.12 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 6.2 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 6.6 and 10.28 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 7.3 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 38.4% of sleep hours? Round your answer to 2 decimal places.

Answer: _____hours

(e)  If 100 visually impaired students were studied, how many students would you expect to have sleep times of more than 10.28 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: ______hours

Answer 1

We are given the distribution here as:

We would be looking at the first 4 parts here:

a) The probability here is computed as:

P(X < = 6.2)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, as:

therefore 10.83% is the required percentage here.

b) The probability here is computed as:

P( 6.6 < X < 10.28)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

therefore 60.80% is the required percentage here.

c) The probability here is computed as:
P(X > 7.3)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

therefore 76.33% is the required percentage here.

d) From standard normal tables, we have:
P(Z < 0.295) = 0.616

Therefore, P(Z > 0.295) = 1 - 0.616 = 0.384

Therefore the value here is computed as:
= Mean + 0.295*Std Dev

= 8.82 + 0.295*2.12

= 9.4454

therefore 9.45 hours is the required time here.