1. The average number of years a person takes to complete a graduate degree program is 3. The standard deviation is 4 months Assume the variable is normally distributed If and individual enrolls in the program, find the probability that it will take:
a. More then 4 years to complete the program
b. Less than 3 years to complete the program
between 3.8 and 4.5 years to complete the program
between 2.5 and 3.1 years to complete the program
#2. The average thickness if books on a library shelf is 8.3 centimeters. The standard deviation is .6 centimeters. If 20% of the books are oversized, fins the minimum thickness of the oversized books on the library shelf. Assume the variable is normally distributed.
Please help the months and % are confusing me.
✅ Answers
Answerer 1
1.
mean = 3
stdev = 1/3 = 0.333333
a)
0.00135
b)
0.5
c)
0.0082
d)
0.5511
2.
mean = 8.3cm
st dev = 0.6cm
the question is confusing
what do they want?
if oversized means greater than the mean, then 50% of the book are oversized
maybe the meaning is
among all the books with thickness greater than 8.3, what is the minimum thickness which gives 20% = 0.2
If so the value is 8.615cm
which means that the number of books with thickness t ranging from 8.3 and 8.615 cm is 20% of the total number of books
I hope this was useful
Answerer 2
1. There are no percentages here. You could give the probabilities as
percentages, but it is not necessary.
Let the number of years taken by a randomly chosen student to complete
be X. Then X has a normal distribution with mean 3 and SD 1/3.
a) The probability that X is greater than 4
=P(X>4)=P(Z>(4-3)/√(1/3)) which standardizes the X,
=P(Z>1.732)
=1-P(Z≤1.732)
=1-0.9584
=0.0416 ≈0.042 to 2DP or 4.2% if you like.
b)P(X<3)=P(Z<0)=0.5
P(3.8<X<4.5)=P((3.8 -3)/√(1/3)<Z<(4.5-3)/√(1/3))
= P( 1.386 < Z<2.598)
=P(Z<2.598) – P(Z<1.386)
=0.9953-0.9186
=0.0767
I leave you to do the last bit.
2. Let the thickness of a randomly chosen book on the shelves
be T cm. You want the smallest value of T, T’ for which
P(T≥T’)=20%=0.2
Standardizing gives P(Z≥(T’-8.3)/0.6)=0.2
So P(Z≤(T’-8.3)/0.6)=0.8
Looking backwards in your tables you get z=0.842
so (T’-8.3)/0.6=0.842 and solve to give the value of T’.