Math 251, Review Quiz For Test 2
Monday, October 27, 2003
Names: ANSWERS .
1. The following represents a group of 200 people attending a Dental Convention.
Has Beard / No Beard / Row TotalMale / 16 / 80 / 96
Female / 0 / 104 / 104
Column Total / 16 / 184 / 200
Let F be the event the attendee is female, M the event the attendee is male, B be the event the attendee has a beard, and N be the event the attendee has no beard.
(a) Compute: P(B), P(B given F), P(B given M).
Ans:P(B) = 16/200 = .08; P(B given F) = 0; P(B given M) = 16/96 = 1/6.
(b) Are the events B and F independent? Are the events B and F mutually exclusive?
Ans:The events B and F are not independent because P(B) P(B given F). Intuitively, they are
not independent because the probability of someone having a beard depends on gender.
The events B and F are mutually exclusive because there are no women at the conference with beards; that is, P(B and F) = 0.
2. (a) President Geraty has recently received permission to excavate the site of an ancient palace. In how many ways can he choose 10 of the 48 graduate students in the School of Religion to join him?
Ans:C48,10 = = 6,540,715,896
(b) Of the 48 graduate students, 25 are female and 23 are male. In how many ways can President Geraty select a group of 10 that consists of 6 females and 4 males?
Ans:C25,6C23,4 = = 177,1008855 = 1,568,220,500
(c) What is the probability that President Geraty would randomly select a group of 10 consisting of 6 females and 4 males?
Ans:The probability is the Answer in (b) Answer in (a) .2398
3. (a) How many different license plates can be made in the form xzz-zzz where x is a digit from 1 to 9, and z is a digit from 0 to 9 or a letter A through Z?
Ans:The number of license plates is 93636363636 = 544,195,584.
(b) What is the probability that a randomly selected license plate will end with the number 00? That is the license plate looks like xzz-z00?
Ans:The number of license plates of the form xzz-z00 is 9363636 = 419,904. Thus the probability a randomly selected license plate is of this form is
419,904 544,195,584 .0007716
4. Consider the random variable x with the following probability distribution.
x 5 7 9 10
p(x).2 .25 .15 .4
(a) Find the mean and standard deviation of x.
Ans:Mean: = 5(.2) + 7(.25) + 9(.15) + 10(.4) = 8.1
Variance: 2 = 25(.2) + 49(.25) + 81(.15) + 100(.4) - 8.12 = 3.79
Thus, the standard deviation is: = 3.791/2 1.947
(b) Sketch a probability distribution histogram for x.
Ans:
5. (a) A physician claims that a certain treatment for prostate cancer is 98% successful at the Hospital where she works. What is the probability that 21 or fewer out of 24 procedures performed at that Hospital will be successful? Assume the 98% success claim is correct and that the results are independent.
Ans:Because the results are independent, this is a binomial experiment with n=24, p=.98 and q=.02. To answer the question, we first compute P(r) for r=24,23,22:
P(24) = (.98)24 0.6158
P(23) = C 24,23 (.98)23(.02)1 240.62830.02 0.3016
P(22) = C 24,22 (.98)22(.02)2 2760.64120.0004 0.0708
Therefore, the probability of 21 or fewer treatments being successful is
1 – P(24) – P(23) – P(22) 1 – 0.9882 = 0.0118
(b) A man seeking treatment decides to investigate this claim, and finds that out of the 24 patients recently treated with this procedure, it was successful on 21 of those patients. Assume results of the treatments are independent. Should he be skeptical of the claimed success rate? Explain.
Ans:He should be skeptical. If the claim were true, there is only a 1.2% chance to see only
21 successes out of 24.
(c) Find the mean and standard deviation for the number of successful treatments out of 24.
Ans:Mean: = (24)(.98) = 23.52
Standard Deviation: = [(24)(.98)(.02)]1/2 .6859
6. The distribution of weights of a type of salmon is normal with a mean of 21 lbs and standard deviation of 3 lbs.
(a) What weight has a percentile rank of 30?
Ans:Approximately 30% of all z-values are to the left of z= -.52, thus z -.52,
thus (x-21)/3 -.52 implies x 19.44
So the 30th percentile is approximately 19.44 lbs.
(b) What proportion of salmon weigh between 15 lbs and 25 lbs?
Ans:P(-2 < z < 1.33) = .9082 - .0228 = .8854
(c) What proportion of salmon weigh less than 15 lbs?
Ans:P(z < -2) = .0228
(d) What proportion of salmon weigh more than 20 lbs?
Ans:P(z > -1/3) = 1 - .3707 = .6293
(e) What proportion of salmon weigh more than 22 lbs?
Ans:P(z > 1/3) = 1 - .6293 = .3707
7. Miscellaneous Questions Regarding Normal Distributions.
(a) Find the z value so that 90% of the normal curve lies between –z and z.
Ans: Find z value so that only 5% of normal curve lies above z (then 5% will lie below –z be symmetry and 90% will be between –z and z). Thus find the z values so that the area to the left of it is .95: z 1.65
(b) Suppose x is a normal random variable with = 50 and = 13.
(i)Convert the interval 37 < x < 48 to a z interval.
Ans: -1 < z < -.154
(ii) Convert the interval x > 71 to a z interval.
Ans:z < 1.615
(iii) Convert the interval z< -1 to an x interval.
Ans:x < 37
(iv) Convert the interval 1 < z < 3 to an x interval.
Ans:63 < z < 89
(c) Let b > 0. If we know that P(-b < z < b) = d, find P(z > b) in terms of d.
Ans: By symmetry, the probability is half of 1-d, i.e., P(z > b) = (1 – d)/2.
8. (From p. 328 #19) Blood type AB is found in only 3% of the population. If 250 people are chosen at random, what is the probability that between 5 and 10 will have this blood type?
Ans:Because np = 7.5 > 5 and nq = 242.5 > 5, we can use the normal approximation to the binomial distribution. Using the continuity correction, we compute
P(4.5 < x < 10.5 ) = P(-1.11 < z < (10.5 – 7.5) 2.69722)
= P(-1.11 < z < 1.11) = .8665 - .1335 = 0.7330
9. (From p. 326 #12) A company determines that the life of the laser beam device in their compact disc player is normally distributed with mean 5000 hours and standard deviation 450 hours. If you wish to make a guarantee so that no more than 5% of the laser beam devices fail during the guarantee period, how many playing hours should the guarantee period cover?
Ans:To answer this, find a z value so that 5% of the standard normal curve is to the left of this z value, and then convert it to an x value:
An appropriate z value is z -1.645, this converts to x = 5000 – 1.645(450) = 4259.75.
So we would guarantee 4260 hours of use.