Homework SourseA student Surveyed 200 people in a statistics lecture to fin
A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15%
B.0.3%
C. 1.5 %
D. 2.5 %
A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15%
B.0.3%
C. 1.5 %
D. 2.5 %
A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15%
B.0.3%
C. 1.5 %
D. 2.5 %
Solution
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 4.55
u = mean = 6.8
s = standard deviation = 0.75
Thus,
z = (x - u) / s = -3
Thus, using a table/technology, the left tailed area of this is
P(z > -3 ) = 0.0015 or 0.15% [OPTION A]
