A recent survey describes the distribution of total sleep time among college stu

Question

A recent survey describes the distribution of total sleep time among college students as approximately Normal with a mean of 6.78 hours and standard deviation of 1.24 hours. You plan to take an SRS of size n-120 and compute the average total sleep time. 3. (a) (2pts) What is the standard deviation for the average time? (b) (2pts) What is the probability that the average time will be 7.5 hours? (c) (2pts) What is the probability that the average time will be more than 8 hours? (d) (2pts) What is the probability that the average time will be below 6.9 hours? In the survey of 4513 college students described in the problem 3,46% of the respondents reported falling asleep in class due to poor sleep. Assume that the students' actions (falling asleep or not) are independent, with the probability of falling asleep equal to 0.46. You randomly sample 1O students in your dormitory. 4. (2pts) What is the probability that 8 students state that they fell asleep in class during the last week due to poor sleep? (a) (b) (2pts) What is the probability that more than 5 students state that they fell asleep in class during the last week due to poor sleep? (c) (2pts) What is the probability that less than 3 students state that they fell asleep in class during the last week due to poor sleep? (d) (2pts) What is the probability that at least 7 students state that they fell asleep in class during the last week due to poor sleep? (3pts) What is the probability that NOT more than 5 students state that they fell asleep in class during the last week due to poor sleep? (e)

Explanation / Answer

Q3.

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 6.78
sample size (n) = 120
a.
standard Deviation ( sd )= 1.24/ Sqrt ( 120 ) =0.1132
b.
P(X > 7.5) = (7.5-6.78)/1.24/ Sqrt ( 120 )
= 0.72/0.113= 6.3606
= P ( Z >6.3606) From Standard Normal Table
= 0
c.
GREATER THAN
P(X > 8) = (8-6.78)/1.24/ Sqrt ( 120 )
= 1.22/0.113= 10.7778
= P ( Z >10.7778) From Standard Normal Table
= 0

d.
P(X < 6.9) = (6.9-6.78)/1.24/ Sqrt ( 120 )
= 0.12/0.1132= 1.0601
= P ( Z <1.0601) From Standard NOrmal Table
= 0.85544

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