1.
0.9987 2.
0.0228 3. 2917.75 4.
3000
5. 0.1587
6. 0.091 7. 3.4657 8. 0.0498
The compressive strength of samples of cement can be modeled by a normal distribution with a mean of
3000 kg/cm 2 and a standard deviation of 50 kg/cm 2 . Answer questions 1 to 5 using this information.
1.
What is the probability that a sample’s strength is less than 3150 kg/cm 2 .
P(X<3150)= P Z

3150

50
3000

P ( Z

3 ) =0.9987
2.
What is the probability that a sample’s strength is between 2800 and 2900 kg/cm 2 .
P(2800

X

2900)= P
2800

50
3000

Z

2900

50
3000

P (

4

Z
 
2 )
=0.0228-0=0.0228
3.
What strength is exceeded by 95% of the samples?
P(X>x*)= P Z
 x *

3000
50

0 .
95 then x *

3000
50
 
1 .
645
If you solve for x*=2917.75, that is the strength exceeded by 95% of the sample.
4.
What is the median strength?
P(X
 median)= P

Z
 median

3000
50

0 .
5 then median
50

3000

0
Median=3000.
Since we already know that normal distribution is symmetric (that means mean=median), we could directly answer without solving it.
5.
What is the probability that strength exceeds the mean strength by more than 1 standard deviations?
P(X>3000+1(50))= P Z

3050

50
3000
=P(Z>1)=0.1587
The time between two successive arrivals at the drive-up window of a local bank is exponentially distributed with the expected time between two successive arrivals of 5 minutes. Answer the next 3 questions using this information.
X ~ Exponential (

=1/

=5) then f(x)=
P(X
 x)= 1
 e

0 .
2 x
P(X>x)= e

0 .
2 x
P(a < X < b) = e

0 .
2 a
e

0 .
2 b
0 .
2 e

0 .
2 x
, x>0.
6.
What is the probability that time until the next arrival is between 9 and 13 minutes?
P(9 < X < 13) = e

0 .
2 ( 9 )
e

0 .
2 ( 13 )
=0.1653-0.0743=0.091
7.
What is the median time between two successive arrivals?
P(X
 median)= 1
 e

0 .
2 median
=0.5 then median=3.4657
8.
What is the probability that time between two successive arrivals exceeds the mean time by more than
2 standard deviations?
P(X>5+2(5))=P(X>15)= e

0 .
2 ( 15 )
=0.0498