Download Answer Key for Quiz 4

Answer key for Quiz 4 (collected the take home quiz on: 2/28/2005 in class)
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/cm2 and a standard deviation of 50 kg/cm2. Answer questions 1 to 5 using this information.
1.
What is the probability that a sample’s strength is less than 3150 kg/cm2.
3150  3000 
  P( Z  3) =0.9987
50



P(X<3150)= P Z 
2.
What is the probability that a sample’s strength is between 2800 and 2900 kg/cm 2.
2900  3000 
 2800  3000
Z
  P(4  Z  2)
50
50


P(2800X2900)= P
3.
=0.0228-0=0.0228
What strength is exceeded by 95% of the samples?


P(X>x*)= P Z 
4.
x * 3000
x * 3000 
 1.645
  0.95 then
50
50

If you solve for x*=2917.75, that is the strength exceeded by 95% of the sample.
What is the median strength?


P(Xmedian)= P Z 
median  3000
median  3000 
0
  0.5 then
50
50

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  3000 
 =P(Z>1)=0.1587
50

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)= 0.2e
P(Xx)= 1  e
P(X>x)= e
0 .2 x
0 .2 a
 0 .2 ( 9 )
0 .2 b
-e
0.2 (13)
=0.1653-0.0743=0.091
What is the median time between two successive arrivals?
P(Xmedian)= 1  e
8.
-e
What is the probability that time until the next arrival is between 9 and 13 minutes?
P(9 < X < 13) = e
7.
, x>0.
0.2 x
P(a < X < b) = e
6.
0.2 x
0.2 median
=0.5 then median=3.4657
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