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GTS 111 Semester 1, 2010
Homework 5
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exam and vice versa. There might be typos/mistakes.
Homework Assignment
Each time at bat the probability that a baseball player gets a hit is 0.300. He comes up to bat four
times in this particular game. Assume that his times at bat are independent trials. Use this information
to answer problem 1-5
(1) If possible, re-write each time at bat of this player as a a Bernoulli trial. Identify the successful
outcomes and p, probability of success of each trial.
Solution: The successful outcome of each trial is the player gets a hit. The probability of success
is p = 0.300
(2) Is this situation a binomial experiment? Explain your reason.
Solution: It is a binomial experiment with fixed number of trials (4 trials). Each trial has two possible
outcomes, a player gets a hit (S) or he didn’t hit the ball(F ) where success is ”get a hit”. The probability
of success is fixed which is 0.3 for each trial. And his times at bat are independent trials.
(3) Find the probability that he only hit the ball once on his third time at bat.
Solution: This question asked for the probability of F F SF .
p(F F SF ) = p(F )p(F )p(S)p(F ) = (0.7)(0.7)(0.3)(0.7)
(4) Find the probability that he gets exactly one hit for the whole game.
Solution: This question asked for the probability of one success.
p( 1 success of 4 trials)
= b(1; 4, 0.3)
= C14 (0.3)1 (0.7)3
(5) Find the probability that he gets at least one hit.
Solution:
p(gets at least one hit)
1
=
1 − p(no hits)
=
1 − b(0; 4, 0.3)
=
1 − C04 (0.3)0 (0.7)4
If the probability of a person contracting influenza on exposure is 0.6, consider the binomial distribution
for a family of 5 that has been exposed. Use this information to answer question 6-10.
(6) Write the distribution function defining this distribution. Also identify ”success” of each trial
Solution: We have p = 0.6 and n = 5. Thus
p(k successes in 5 trials) = Ck5 (0.6)k (0.4)5−k
(7) Construct a table for this binomial distribution.
Solution:
value of random variable, k
p(k successes)
0
C05 (0.6)0 (0.4)5
1
C15 (0.6)1 (0.4)4
2
C25 (0.6)2 (0.4)3
3
C35 (0.6)3 (0.4)2
4
C45 (0.6)4 (0.4)1
5
C55 (0.6)5 (0.4)0
(8) Draw a histogram for this binomial distribution.
(9) Find the mean and the standard deviation for this distribution.
Solution: n = 5, p = 0.6
Mean is 5(0.6) = 3
Variance is 5(0.6)(0.4) = 1.2
p
Standard Deviation is 5(0.6)(0.4) = 1.09
(10) Find the probability that the number of successes lies within 1 standard deviation of the mean.
Solution: Since the mean of the successes is 3. Therefore the number of successes lies within 1 standard
deviation of the means are 3 ± 1.09 = 1.91 − 4.09 or k = 2, 3, 4. Thus the probability that the number of
successes lies within 1 standard deviation of the mean is
p(2 successes in 5 trials) + p(3 successes in 5 trials)p(4 successes in 5 trials)
= b(2; 5, 0.6) + b(3; 5, 0.6) + b(4; 5, 0.6)
= C25 (0.6)2 (0.4)3 + C35 (0.6)3 (0.4)2 + C45 (0.6)4 (0.4)2
2