Math153                                              Test2 Review (Blue Practice Test)

Spring2005 (11/99)

 

1)      Check all numbers which can be probabilities.

 

____ 2.01              _____ 2/7                    _____ 0

 

____ 1.01              _____ -0.12                 _____ 0.01

 

2)      A lotto game consists of choosing 6 numbers from the numbers 1, 2, … 50.

a)      Find the number of ways of choosing 6 numbers from a set of 50 numbers.

 

 

 

b)      What is the probability of winning the game if two sets of 6 numbers are selected?

No E notation.

 

 

 

3)      Assume that the probability that a taxpayer correctly completes the federal tax form is 70%.  Assume 10 taxpayer are randomly chosen.  Round off to three decimal places.

 

a)      what is the probability that exactly5 complete the form correctly? _______

 

 

b)      what is the probability that 5 or fewer complete the form correctly? ___________

 

 

 

 

 

c)      what is the probability that more than 5 complete the form correctly? ___________

 

 

 

 

 

4)      The probability that an o-ring holds in a rocket is 0.97.  Assume that the o-rings hold independently of each other, and that a rocket has six o-rings. 

What is the probability that all 6 o-rings hold (and the mission is successful)?

 

                                                                                                                  ___________


5)  Assume that 85% of people who start high school graduate from high school.

Let x = number of people who graduate from high school out of 1000 randomly selected people who began high school.

a)      Find the mean and standard deviation of x.  (Round off answers to two decimal places.)

m    =  ___________

 

s    =  ___________

 

 

b)      In a certain community, 1000 people are randomly selected, and 800 who began high school also graduated from high school.

Find the z-score for this value of x.  ____________

 

 

 

 

Is this community unusual? ____ You MUST briefly explain

 

 

 

 

 

 

 

 

6)  Find the following probabilities for a random variable, z, which has the standard normal distribution.  Round off your answers to four decimal places.

a)      P(-1.36 < z < 1.07) = ____________________

 

 

 

b)      P( z < -0.23) = ____________________

 

 

 

 

-CONTINUED-


7.  Assume that the cholesterol levels for a certain population of vegetarians are normally distributed with a mean of 162 and a standard deviation of  15.6. 

a)      What is the probability that the cholesterol level for a randomly chosen person from this population is between 150 and 175? (First find the z-score(s) to two decimal places.  Secondly, round off answer to 4 decimal places.)

z-score(s): ________

answer: __________

 

 

 

 

b)      What is the probability of a randomly chosen person from this population having a cholesterol level greater than 200? (First find the z-score(s) to two decimal places.  Secondly, round off answer to 4 decimal places.)

z-score(s): _________

 

answer: ___________

 

 

 

 

 

 

 

 

8)     The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.  If we stipulate that a baby is premature if the length of pregnancy is in the lowest 4%, find the length (in days) that separates premature babies from those who are not premature.  Round off to the nearst whole number of days.                                           

                                                                                                                        ______________

                                                                                                                                   


Answer Key                 Test2Review

3/05

 

1.)  2/7, 0, 0.01

 

2) a) 50 C 6 = 15, 890, 700     b) 2/15, 890, 700 or 1.258597796x10-7 

 

3) n=10, success: completes correctly, p = .7

a) .103

b) .103+.037+.009 +.001 + 0 + 0 = .15

c) .200 + ..267 + .233 + .121 + ..028 = .849

 

4) n= 6, success:O-ring holds, p=.97

  P(x=6) = 6 C 6 (.97)6  (.03)0 = .83 or 83%

 

5) m = np = 1000 (.85) = 850

     s  =  =  = 11.29

z = x - m  = 800-850 = -4.43

          s             11.29

Yes, this z-score is unusual because it is less than –2.

 

6) a) .7708       b) .4090

 

7) a)  z = (150-162)/15.6 = -.77 and z = (175-162)/15.6 =.83

normalcdf(-.77, .83) = .5761

 

b)  z = (200-162)/15.6 = 2.44

normalcdf(2.44, 10) = .0073

 

8)  invNorm(.04) = -1.75

z = x- m

        s 

 

-1.75 = x – 268

                   15

x = 241.75 or 242 days