Question 12.
In Exercises 12, determine whether the random variable X has a binomial distribution. If it does, state the number of trials n . If it does not, explain why not . Ten students are chosen from a statistics class of 300 students. Let X be the number who got an A in the class.
The random variable X does not have a binomial distribution because the probability of each trial is not known and thus it is not known if it is the same or not.
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Question: 60)
Boys and girls: A couple plans to have children until a girl is born, but they will have no more than three children. Assume that each child is equally likely to be a boy or a girl. Let Y be the number of boys they have.
Find the probability distribution of X .
We need to get the probability distribution of X.
The random variable X is defined as the number of girls the couple have. It is stated that the couple will not have more than three children.
Therefore, X can be 0, 1, 2 or 3.
Besides, each child has equal chance of being a girl or a boy. That is, probability of a boy is 0.5 and that of a girls is 0.5.
For X = 0, the couple does not have boys, meaning that they will have girl. Because the couple plans to have children until they get a girl, the probability is 0.5.
For X = 1, the couple can give birth to one boy. Because the couple plans to have children until they get a girl, the second child is a girl.
Therefore, P (1) = GB
= 0.5 X 0.5
= 0.25
Therefore, the probability of the couple having only one boy is 0.25.
For X = 3, the couple may have three boys. It is stated that the couple will not have more than three children.
Therefore, P (3) = GGB
= 0.5 x 0.5 x 0.5
= 0.125
Therefore, the probability of the couple having three boys is 0.125.
For X = 3, the couple may have three boys. It is stated that the couple will not have more than three children.
Therefore, P (3) = GGG
= 0.5 x 0.5 x 0.5
= 0.125
Therefore, the probability of the couple having three boys is 0.125.
Thus, the probability distribution of X can be represented in a table as follows:
| X | Child | P (X) |
| 0 | G | 0.5 |
| 1 | GB | 0.25 |
| 2 | GGB | 0.125 |
| 3 | GGG | 0.125 |
Find the mean μ X
To get the mean μx, the general for mean of a discrete random variable Y is
μy = ∑[yP(y)]
Thus, the mean (μx) is,
μx = ∑ [xP(x)]
n= 3
x = 0
= (0 x 0.5) + (1 x 0.25) + (2 x 0.125) + (3 x 0.125)
= 0 + 0.25 + 0.25 + 0.375
= 0.875
Thus, the mean μX = 0.875
Find the standard deviation σ X .
To get the standard deviation σ X , the general formula for variance and standard deviation of a discrete random variable x is
= [(0 x 0.5) + (12 x 0.125) + (22 x 0.125) + (32 x 0.125)] – (0.875)2
= [(0x0.5) + (1 x 0.125) + (4 x 0.125) + (9 x 0.125)] – (0.875)2
= [0 + 0.25 + 0.5 + 1.125] – 0.7656
= 1.875 – 0.7656
= 1.1094
The standard deviation σX is,
σX = σX
=
= 1.0533
Thus, the standard deviation σX = 1.0533.
Question 30
Baby weights: The weight of male babies less than 2 months old in the United States is normally distributed with mean 11.5 pounds and standard deviation 2.7 pounds
30) A. Find the 81st percentile of the baby weights
First, we determine how many standard deviations above the mean one could have to be in the 81 st percentile. It is gotten by using the z table and getting the z associated with 0.81, which is 0.7910. Therefore, one should 0.7910 standard deviations above the mean to be in the 81 st percentile.
Second, the standard deviation is 2.7, one should be:
(2.7)(0.7910) = 2.1357
Thus, the 81 st percentile = 11.5 + 2.1357 = 13.6357
B. Find the 10th percentile of the baby weights.
Using the formula in part (a) above, the z value for 10 th percentile, that is, 0.1 = 0.5398.
The standard deviation = (2.7) x (0.5398) = 1.45746
Therefore, the 10 th percentile = 11.5 + 1.45746 = 12.95746
Watch your cholesterol: The National Health and Nutrition Examination Survey (NHANES) reported that in a recent year, the mean serum cholesterol level for U.S. adults was 202, with a standard deviation of 41 (the units are milligrams per deciliter). A simple random sample of 110 adults is chosen.
22) A. What is the probability that the sample mean cholesterol level is greater than 210?
Mean = 202
Standard Deviation = 41
Sample size = 110
Standard deviation of sample mean = 41 110 = 3.9
Z value for 210 is (210-202)/3.9 = 2.0465. The p value for 2.0465 is 0.9796
P (X<210) = 0.9796
The probability that the sample mean cholesterol level is greater than 210 = 1 – 0.9796 = 0.0204.
Question 32
Find the levels of the confidence intervals that have the given critical values
32) 1.04
From the z table, the confidence interval of 1.04 is 0.70.
Question 28
Baby talk: In a sample of 77 children, the mean age at which they first began to combine words was 16.51 months, with a standard deviation of 9.59 months
Construct a 95% confidence interval for the mean age at which children first begin to combine words.
First, we need to compute t-critical value, and then get the confidence interval.
The t-critical value for 95% confidence level is α = 0.05
The sample size is small and two-tailed test. Thus, from the t distribution table, we look in the column headed and the row headed by using the degree of freedom, which is,
d.f. = n -1
= 77-1
= 76
The t critical value for the 95% confidence interval = 1.992
95% Confidence Interval = x̃ ± t c x s/√n
= 16.51 ± 1.992x9.59/√77
= 16.51 ± 2.18
= 14.33 to 18.69
A 95% confidence interval for the mean age at which children first start combining words = 14.33 – 18.69.
If a sample of 50 children had been studied, would you expect the confidence interval to be wider or narrower than the interval constructed in part (a)? Explain.
For a sample of 50, the t-critical value for 95% confidence level is α = 0.05
d.f. = n -1
= 50-1
= 49
The t-critical value for the 95% confidence interval = 2.010
95% Confidence Interval = x̃ ± t c x s/√n
= 16.51 ± 2.010x9.59/√50
= 16.51 ± 2.73
= 13.78 to 19.24
A 95% confidence interval for the mean age at which children first start combining words = 13.78 to 19.24.
Therefore, if a sample of 50 children had been studied, the confidence interval would be expected to be wider than that of part a, since the sample size reduces from 77 to 50.
Question:
10) In a simple random sample of size 60, there were 38 individuals in the category of interest
A) Compute the sample proportion
=
=
= 0.63
B) Assume the assumptions for a hypothesis test satisfied
Assumptions
np ≥ 10, n (1- p) ≥ 10
The assumption is approximate by a normal distribution
n = 60
np = 60 x 0.70
= 42
n(1- p) = 60 x (1 – 0.70)
= 18
np ≥ 10, n (1- p) ≥ 10
The assumption is approximate by a normal distribution n = 60 is satisfied.
C) It is desired to test H 0 : p = 0.7 versus H 1 : p ≠ 0.7. Compute the test statistic z .
To test the hypothesis that the population proportion differs from 0.70 at 5% significant level.
The null and alternative hypothesis is:
H 0 : P = 0.70
H 1 : P ≠ 0.70
Using MINITAB, the z-test is found to be:
Test of p = 0.7 vs. p not = 0.7
Sample = 1
X = 38
N = 60
Sample p = 0.633333
95% C I = (0.511399, 0.755267)
Z-value = -1.13
P-value = 0.260
D) Do you reject H 0 at the 0.05 level?
The null hypothesis will not be rejected at 0.05 significance level, because the p-value of 0.260 is higher than 0.05. There is inadequate evidence to show that the population proportion differs from 0.70. The outcome is statistically insignificant.
Question 10 section 9.7
A test has power 0.80 when μ 1 = 3.5. True or false:
C) The probability of making a Type I error when μ 1 = 3.5 is 0.20.
The answer is true. This is because the significance level is 0.80 for Type I error. At the significance level of 0.8, it means that the null hypothesis of 0.8 is true.
D) The probability of making a Type II error when μ 1 = 3.5 is 0.20.
The answer is true. This is because the probability of making Type II error when alternative hypothezed mean 3.5 is 0.20 because when 0.80 is subtracted from 1 the answer is 0.20. The null hypothesis at 0.2 will be rejected. Thus, the statement is true.
