Download Math 1342 Review 1

Math 1342 Review 3(answers)
1. The probability density function for a continuous random variable X, which takes on values
from 1 to 7 inclusive, is given by the following graph.
1
3
1
4
7
a) Verify that the probability density function is valid by showing that the total area under
its graph is 1.
1
1
The function is nonnegative, and area   6   1.
2
3
b) Find P  X  2 
1 1
1 17
P  X  2  1  P  X  2   1  1  1  
2 9
18 18
c) Find P  2  X  6 
1 1 1 1
1 8
P  2  X  6  1  P  X  2  P  X  6   1  1  1  1  
2 9 2 9
9 9
d) Using the idea of a balance point, what’s the mean or expected value of X?
4
e) Using the idea that the median separates the upper 50% of the values from the lower 50%
of the values, what’s the median of X?
4
f) Using the idea that the mode is the most likely value to occur, what’s the mode of X?
4
2. If Z is a standard normal random variable, then find the following probabilities:
a) P  Z  1.32 
b) P  2.5  Z  1.5
.9066
c) P  Z  1.53
 P  Z  1.53  .0630
P  Z  1.5  P  Z  2.5  .0668  .0062  .0606
d) P  Z  1.75
 P  Z  1.75  .9599
3. If X is normal with mean of 210 and standard deviation of 20, then find the following
probabilities:
a) P 160  X  220 
220  210 
 160  210
P
Z 
  P  2.5  Z  .5  .6915  .0062  .6853
20
20


b) P  X  150 
150  210 

P Z 
  P  Z  3  .0013
20


c) P  X  240 
240  210 

P Z 
  P  Z  1.5  .0668
20


d) P  X  200 
200  210 

P Z 
  P  Z  .5  .6915
20


4. Find the value of c if
a) P  c  Z  0   .3264
P  Z  c   .5  .3264  .1736  c  .94
b) P  Z  c   .1515
P  Z  c   .1515  c  1.03  c  1.03
c) P  Z  c   .5
P  Z  c   .5  c  0
d) P  c  Z  c   .8502
P  Z  c   .0749  c  1.44  c  1.44
5. A normal random variable has a standard deviation of 20. Find its mean if the probability is
.2946 that it will take on a value less than 80.
80   
80  

P  X  80   .2946  P  Z 
 .54    90.8
  .2946 
20 
20

6. The lengths of sardines are normally distributed with mean of 4.54 inches and standard
deviation of .26 inches. What percentage of sardines are
a) less than 4 inches?
4  4.54 

P  X  4  P  Z 
  P  Z  2.08   .0188  1.88%
.26 

b) between 4.4 and 4.8 inches?
4.8  4.54 
 4.4  4.54
P
Z 
  P  .54  Z  1  .8413  .2946  .5467  54.67%
.26
 .26

7. A virus is believed to infect 2% of the population. If a sample of 3,000 people are selected
at random and tested, use a normal random variable to estimate the probability of the
following:
a) less than 30 are infected
29.5  60 

P  X  30   P Xˆ  29.5  P  Z 
  P  Z  3.98  .000034
58.8


b) between 40 and 80, inclusive, are infected
80.5  60 
 39.5  60
P  40  X  80   P 39.5  Xˆ  80.5  P 
Z
  P  2.67  Z  2.67 
58.8
58.8






 .9962  .0038  .9924
c) at least 70 are infected
69.5  60 

P  X  70   P Xˆ  69.5  P  Z 
  P  Z  1.24   .1075
58.8


8. A normally distributed population has a mean,  , of 24 and a standard deviation,  , of 12.
If a random sample of size 36 is selected and the sample mean, X , is computed, then
a) Determine the mean of the sample means,  X .


24
b) Determine the standard deviation of the sample means,  X .
12

 2
36
c) Determine P X  26 .


26  24 

P  X  26   P  Z 
  P  Z  1  .1587
2 

d) Determine P  X  22.5 .
22.5  24 

P  X  22.5  P  Z 
  P  Z  .75  .2266
2


e) Determine P  23  X  27  .
27  24 
 23  24
P  23  X  27   P 
Z 
  P  .5  Z  1.5  .9332  .3085  .6247
2 
 2
9. Here is the frequency distribution for a population
X frequency relative frequency
1
0
4
4
3
1
12
4
Here is the probability distribution for the population
The population mean,  , is 34 , and the population standard deviation,  , is
samples of size 2(without duplicates), then
a) complete the probability distribution of the sample means.
P X
X
3
4
. If we take
 
0
C2
1

C 20
16 2
.5
4  12 2

C
5
16 2
1
4
C2 11

C 20
16 2
12
b) Calculate the mean of the sample means,  X , and verify that it’s the same as the
population mean.
1 1 2
11 15 3
 X  0     1    , the same as the population mean.
20 2 5
20 20 4
c) Calculate the standard deviation of the sample means,  X , and verify that it’s

approximately equal to
.
2
9 1
1 2 1 11
9
1
11
7
X 
    




 .295803898,
16 20 16 5 16 20
320 40 320
80
X
2

3
 .306186217,so they're pretty close.
4 2
10. A random sample of size n  200 is obtained from a population whose size is N  25,000
and whose population proportion of a characteristic is p  .65 .
a) What’s the mean of the sample proportions,  p ?
.65
b) What’s the standard deviation of the sample proportions,  p ?



.65 1  .65 
 .033726843
200
c) Determine P p  .68 .


P p  .68  P  Z 








.68  .65 
 P  Z  .89   .1867

.65.35

200


d) Determine P p  .59 .


P p  .59  P  Z 







.59  .65 
 P  Z  1.78  .0375

.65.35

200
