Question

1/ Compute the standard deviation for ages of British nurses in 1851. Assume that the table below shows the age distribution of nurses in Great Britain in 1851. Round your answer to nearest hundredth. Age range (yr) 20-29 30-39 40-49 50-59 60-69 70-79 80+ Midpoint (x) 24.5 34.5 44.5 54.5 64.5 75.5 84.5 Percent of nurses 5.8% 9.8% 19.6% 29.1% 24.9% 9.0% 1.8% Select one: a. 53.76 b. 17.25 c. 1.53 d. 3.27 e. 13.69

2. Richard has been given a 12-question multiple-choice quiz in his history class. Each question has four answers, of which only one is correct. Since Richard has not attended the class recently, he doesn't know any of the answers. The success occurs if Richard answers a question correctly and the failure occurs if Richard is unable to answer a question correctly. Assuming that Richard guesses on all 12 questions, find the probability that he will answer no more than 3 questions correctly. Round your answer to the nearest thousandth. Select one: a. 0.258 b. 0.649 c. 0.25 d. 0.063 e. 0.5

3.The probability of a radar station detecting an enemy plane is 0.65 and the probability of not detecting an enemy plane is 0.35. If 100 stations are in use, what is the expected number of stations that will detect an enemy plane? Select one: a. 65 b. 100 c. 98 d. 0 e. none of these choice

4. The probability of a single radar station detecting an enemy plane is 0.6. How many such stations are required to be 97% certain that an enemy plane flying over will be detected by at least one station? Select one: a. 5 b. none of these choices c. 2 d. 3 e. 4

5.

There are 4 radar stations and the probability of a single radar station detecting an enemy plane is 0.55. Make a histogram for the probability distribution.

r
0	0.041
1	0.200
2	0.368
3	0.300
4	0.092

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Answer 1

(1) Here, x = (24.5, 34.5, 44.5, 54.5, 64.5, 75.5, 84.5) and probability = (5.8%, 9.8%, 19.6%, 29.1%, 24.9%, 9.0%, 1.8%).
standard deviation = 13.69.

(2) X ~ Binomial(n = 12, p = 0.25). To find, P(X <= 3) = 0.649.

(3) Expected number = 100 * 0.65 = 65.

(4) X ~ Binomial(n, p = 0.6). To find 'n' such that P(X >= 1) = 0.97.
By trial and error, we find n = 4.

(5)