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timer Asked: Jun 24th, 2018
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Please see file attached. I will send the data. I need it within 1 hour. Please help.................

Homework 2 Due Jun 22 1. Each child born to a particular set of parents has a probability 0.25 of having blood type O. If these parents have 3 children. Define the variable X to be the number of blood type O in the children. a) Find the probability model (sample space and probability) for the variable. b) What is probability that at least 1 child have blood type O. c) What is the mean and standard deviation of X? 2. If we define X to be the number of bad checks received on a randomly selected day by a store, X has the respective probabilities as the following. X 0 Pr(X) 0.05 1 0.1 2 0.2 3 0.4 4 0.15 5 0.1 a). Find the population mean and variance of X. b). Define a random variable L to be the loss of the store, L= -100X+2. Find the mean and variance of L. c). If Y is the number of bad checks received on a randomly selected day by a different store which is dependent on the first store, i.e., Y and X are dependent. Specifically, Y=0 if X < 4, and Y=1 otherwise What is the probability model of Y? What is the mean and variance of Y? 3. Suppose there are two variables X and Y, which are independent and follow an identical probability model: X 0 Pr(X) 0.2 1 0.5 2 0.3 Y 0 Pr(Y) 0.2 1 0.5 2 0.3 a) What is the mean and variance of X+Y? Hint: use the property Var(X+Y)=Var(X)+Var(Y), but note that this property works only when X and Y are independent. b) What is the mean and variance of X-Y? c) What is the mean, variance, and standard deviation of (X+Y) 2 4. A certain drug treatment cures 90% of cases of hookworm in children. Suppose that 10 randomly selected children suffering from hookworm are to be treated, and that the children can be regarded as a random sample from the population. Find the probability that a) no more than 80% will be cured b) no less than 85% and no more than 95% will be cured. c) What is the mean and standard deviation of the number of cured out of 10? d) [PQ] Suppose 100 children are randomly chosen, how would the answer to question a) and b) change? What the relationship do you observe between the sample size and the answer? (Hint: use JMP to compute the probability with Binomial distribution). 5. The Shell of the land snail Limocolaria martensiana has two possible color forms: streaked and pallid. In a certain population of these snails, 60% of the individuals have streaked shells. Suppose that a random sample of 10 snails is to be chosen from this population. Find the probability that the percentage of streaked-shelled snails in the sample will be a) 50%. b) 60% c) 70% d) What is the mean number of streaked-shelled snails? e) What is the standard deviation of the number of streaked-shelled snails? 6. Heights of a certain population of com plants follow a normal distribution with mean 145 cm and standard deviation 22 cm. What percentage of the plant eights are (a) 120 cm or less? (b) Between 120 and 150 cm? (c) Between 130 and 180 cm? (d) Given the condition that it is between 120cm and 150cm, what is the probability that is between 130cm and 180cm? (e) What is the 95th and 90th percentile of the distribution?

Tutor Answer

Neginn
School: New York University

Attached.

Homework 2 Due Jun 22
1. Each child born to a particular set of parents has a probability 0.25 of having blood type
O. If these parents have 3 children. Define the variable X to be the number of blood type
O in the children.
a) Find the probability model (sample space and probability) for the variable.
X
Pr(X)

0
0.4219

1
0.4218

2
0.1407

3
0.0156

Pr(0) = (30) × 0.753 × 0.250 = 0.4219
Pr(1) = (31) × 0.752 × 0.251 = 0.4218
Pr(2) = (32) × 0.751 × 0.252 = 0.1407
Pr(3) = (33) × 0.750 × 0.253 = 0.0156
b) What is probability that at least 1 child have blood type O.
Pr(x ≥ 1) = 1 – Pr(0) = 1 – 0.4219 = 0.5781
c) What is the mean and standard deviation of X?
Mean = np = 0.25*3 = 0.75
Var(X) = np(1 - p) = 3*0.25*0.75 = 0.5625
SD(X) = (Var(X))0.5 = 0.75
2. If we define X to be the number of bad checks received on a randomly selected day by a
store, X has the respective probabilities as the following.
X
0
Pr(X) 0.05

1
0.1

2
0.2

3
0.4

4
0.15

5
0.1

a). Find the population mean and variance of X.
Mean = 0*P(0) + 1*P(1) + 2*P(2) + … + 5*P(5) = 0 + 0.1 + 0.4 +1.2 + 0.6 + 0.5 = 2.8
Var(X) = (0-2.8...

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Anonymous
Outstanding Job!!!!

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