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Probability Distributions
The variables in problem 5.1 are either discrete or continuous. Which are they and why?
Number of siblings -- continuous. This is because the number of siblings cannot change at the moment
Number of conversations -- discrete. There will be more conversations to be held between the children and the mothers.
Explain why the variable "number of dinner guests for Thanksgiving dinner" is discrete.
The variable is discrete because the number of guests can change through cancelation or extra invites.
Explain why the variable "number of miles to your grandmother's house" is continuous.
The variable is continuous because the distance cannot change and the house will not move as it is stationary.
Problem
Express P (x) =1/6; for x = 1, 2, 3, 4, 5, 6, in distribution form.
P (1) = 1/6
P (2) = 2/6 = 1/3
P (3) = 3/6 = 1/2
P (4) = 4/6 = 2/3
P (5) = 5/6
P (6) = 6/6
b. Construct a histogram of the probability distribution P (x) =1/6; for x = 1, 2, 3, 4, 5, 6, in distribution form
1/6
1/6
1/6
4/6
4/6
6/6
c. Describe the shape of the histogram in part (b)
Dog food distribution
Problem 5.18
Find the mean and standard deviation of the number of ships that arrive at a harbor on a given day
x
P (x)
xP (x)
x2
x2 P (x)
10
0.4
4
40
11
0.2
2.2
24.2
12
0.2
2.4
28.8
13
0.1
1.3
16.9
14
0.1
1.2
19.6
60
1
11.1
Mean = ? = ?[xP (x)] = 11.1
Variance = 129.5 -- (11.1)2 = 6.29
Standard Deviation = ?6.29 = 2.508
Problem 5.28
State a very practical reason why the defective item in an industrial situation might be defined to be the "success" in a binomial experiment
Supposing this industrial unit obtained numerous complaints and criticisms. The supervision representatives are coming in to ascertain these criticisms. Taking this into consideration, checking and proving the items and ascertaining them to be defective can be well-defined as 'success' for the government supervision representatives.
Problem 5.48
Consider the binomial distribution where n = 11 and p = 0.05 (see problem 5.47).
a. Use the distribution [problem 5.47(b) or Table 2] and find the mean and standard deviation using formulas (5.1), (5.3a), and (5.4).
Using formula 5.7 and 5.8,
the mean in np = 0.55
and the standard deviation is ?npq
=?11*0.05*0.95 = 0.7228
b. Compare the results of part (a) with the answers found in problem 5.47(a).
The results would have been the same if we had used formulas (5.1), (5.3a), and (5.4).
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