YOUR SIGNATURE: equations is 134 minutes with a standard deviation of 16. Assume
ID: 3323534 • Letter: Y
Question
YOUR SIGNATURE: equations is 134 minutes with a standard deviation of 16. Assume the time he takes to finish an assignment is normally distributed (a) (4 pts) what are the intervals that mark the middle 68%? (8) The time in minutes it takes Gio to finish a homework asignment in differential (b) (4 pts) What is the probability that it takes Gio less than 150 minutes? (c) (4 pts) Ten percent of his longest assigments take longer than how many minutes? (d) (4 pts) What is the probability that Gio misses this class?Explanation / Answer
NORMAL DISTRIBUTION
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 ~ N(0,1)
mean ( u ) = 134
standard Deviation ( sd )= 16
a.
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [134 ± 16]
= [ 134 - 16 , 134 + 16]
= [ 118 , 150 ]
b.
P(X < 150) = (150-134)/16
= 16/16= 1
= P ( Z <1) From Standard Normal Table
= 0.8413
c.
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.281552
P( x-u / (s.d) > x - 134/16) = 0.1
That is, ( x - 134/16) = 1.281552
--> x = 1.281552 * 16+134 = 154.504825 ~ 155
d.
could not able to solve the answer since specfic time interval is not
mentioned to solve on missing the class
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