You know the value of expectation, 7.8h, you know the deviation, 0.75h and you know that the amount of students, sleeping more LESS than (or equal) 9 hours is:
/bigPhi((9h-7.8h)/0.75h)=
/bigPhi(1.6)
Hence the required probability to find such a student (sleeping MORE than 9 hours) is:
P=1-/bigPhi(1.6)
College students average 7.8 hours of sleep per night with a standard deviation of 45 minutes. If the amount of sleep is normally distributed, what proportion of college students sleep for more than 9 hours?
3 answers
Z = (x-μ)/SD
Z = (9-7.8)/.75 = 1.6
In table in the back of your stat text called something like "areas under normal distribution," look up Z score for smaller area to get proportion.
I hope this helps a little more.
Z = (9-7.8)/.75 = 1.6
In table in the back of your stat text called something like "areas under normal distribution," look up Z score for smaller area to get proportion.
I hope this helps a little more.
;kjl
1 answer