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Algebra and Trigonometry (MindTap Course List)

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter14: Counting And Probability

Section14.3: Binomial Probability

Problem 37E

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Question

please use the Statistical tables provide!!

Probability
Poisson distribution
P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = P(A) x P(B) for independent events
X- Po(A)
P(A and B) = P(A) x P(B|A) for dependent events
P(X - x) = **
x!
P(A and B)
P(B|A) =
P(A)
Confidence Interval
Normal Distribution
1. z-confidence
interval:
X- N(4, ơ)
X-u
2. t-confidence
interval:
1. Standard normal: Z -
(df -n-1)
3. Confidence interval
for proportion:
P(1 – P)
n
Sampling Distribution
1. 4 =4
2.
Test Statistics
X-4
1. z-test for p : 2=
3.
X - µ
2.
t-test for u: t =
(d.f. = n-1)
S/ Vn
Discrete Probability Distribution
3. z-test forp: z-
» np 2 5
1. EÇX) = µ, = Ex,P,
P(1 - P),
and np z 5 (where q = 1-p)
Var(X) = o = 2(x, -4,)°p,
%3D
%3D
4. Chi-square test statistic = y0- E)
(0-E)
Binomial Distribution
Σ
X- Bin(n, p)
n-k
P(X - k) -
P" (1-
n!
p (1-
n-k
k!(n -k)!

Question 1
A sports shop selects a random sample and records the sales of items made in China and those
made in America.
Country of Origin
China
Mean
22.5
Std. Deviation
9.75
50
America
33
33.15
11.6
a) What is the SE Mean for the sales from China?
b) Calculate a 95% confidence interval for the mean sales amount made in America

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Population Genetics In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypesthat is, instances in which the two alleles carried at a particular site on an individuals chromosomes are both the same. For population in which blood-related individual mate, them is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. in general. the frequency of homozygous children from mating of blood-related parents is greater than that for children from unrelated parents Measured over a large number of generations, the proportion of heterozygous genotypesthat is, nonhomozygous genotypeschanges by a constant factor 1 from generation to generation. The factor 1 is a number between 0 and 1. If 1=0.75, for example then the proportion of heterozygous individuals in the population decreases by 25 in each generation In this case, after 10 generations, the proportion of heterozygous individuals in the population decreases by 94.37, since 0.7510=0.0563, or 5.63. In other words, 94.37 of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For mating between siblings 1 can be determined as the largest value of for which 2=12+14. This equation comes from carefully accounting for the genotypes for the present generation the 2 term in terms of those previous two generations represented by for the parents generation and by the constant term of the grandparents generation. a Find both solutions to the quadratic equation above and identify which is 1 use a horizontal span of 1 to 1 in this exercise and the following exercise. b After 5 generations, what proportion of the population will be homozygous? c After 20 generations, what proportion of the population will be homozygous?
Dividing a Jackpot A game between two pIayers consists of tossing coin. Player A gets a point if the coin shows heads, and player B gets a point if it shows tails. The first player to get six points wins an $8000 jackpot. As it happens, the police raid the place when player A has five points and B has three points. After everyone has calmed down, how should the jackpot be divided between the two players? In other words, what is the probability of A winning (and that of B winning) if the game were to continue? The French mathematicians Pascal and Fermat corresponded about this problem, and both came to the same correct conclusion (though by very different reasoning's). Their friend Roberval disagreed with both of them. He argued that player A has probability of Winning, because the game can end in the four ways H, TH, TTH, TTT, and in three of these, A wins. Roberval’s reasoning was wrong. Continue the game from the point at which it was interrupted, using either a coin or a modeling program. Perform this experiment 80 or more times, and estimate the probability that player A wins. Calculate the probability that player A wins. Compare with your estimate from part (a).

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