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Chapter 5: Problem 16

Find the mean and standard deviation for a binomial distribution with these values: a. \(n=1000, p=.3\) b. \(n=400, p=.01\) c. \(n=500, p=.5\) d. \(n=1600, p=.8\)

Short Answer

Expert verified
Answer: a. Mean: 300, Standard Deviation: 14.49 b. Mean: 4, Standard Deviation: 1.99 c. Mean: 250, Standard Deviation: 11.18 d. Mean: 1280, Standard Deviation: 16

Step by step solution

01

Find the mean (渭)

Use the formula 渭 = \(n \times p\). Mean (渭) = \(1000 \times 0.3 = 300\)
02

Find the standard deviation (蟽)

Use the formula 蟽 = \(\sqrt{n \times p \times (1-p)}\). Standard Deviation (蟽) = \(\sqrt{1000 \times 0.3 \times (1-0.3)} = \sqrt{1000 \times 0.3 \times 0.7} = \sqrt{210} \approx 14.49\) #b. n=400, p=.01#
03

Find the mean (渭)

Use the formula 渭 = \(n \times p\). Mean (渭) = \(400 \times 0.01 = 4\)
04

Find the standard deviation (蟽)

Use the formula 蟽 = \(\sqrt{n \times p \times (1-p)}\). Standard Deviation (蟽) = \(\sqrt{400 \times 0.01 \times (1-0.01)} = \sqrt{400 \times 0.01 \times 0.99} = \sqrt{3.96} \approx 1.99\) #c. n=500, p=.5#
05

Find the mean (渭)

Use the formula 渭 = \(n \times p\). Mean (渭) = \(500 \times 0.5 = 250\)
06

Find the standard deviation (蟽)

Use the formula 蟽 = \(\sqrt{n \times p \times (1-p)}\). Standard Deviation (蟽) = \(\sqrt{500 \times 0.5 \times (1-0.5)} = \sqrt{500 \times 0.5 \times 0.5} = \sqrt{125} \approx 11.18\) #d. n=1600, p=.8#
07

Find the mean (渭)

Use the formula 渭 = \(n \times p\). Mean (渭) = \(1600 \times 0.8 = 1280\)
08

Find the standard deviation (蟽)

Use the formula 蟽 = \(\sqrt{n \times p \times (1-p)}\). Standard Deviation (蟽) = \(\sqrt{1600 \times 0.8 \times (1-0.8)} = \sqrt{1600 \times 0.8 \times 0.2} = \sqrt{256} = 16\) Thus, we have calculated the mean and standard deviation for each of the given binomial distributions: a. Mean: 300, Standard Deviation: 14.49 b. Mean: 4, Standard Deviation: 1.99 c. Mean: 250, Standard Deviation: 11.18 d. Mean: 1280, Standard Deviation: 16

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Most popular questions from this chapter

Fast Food and Gas Stations Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of \(n=25\) Americans who travel by car are asked how they determine where to stop for food and gas. Let \(x\) be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway. a. What are the mean and variance of \(x ?\) b. Calculate the interval \(\mu \pm 2 \sigma\). What values of the binomial random variable \(x\) fall into this interval? c. Find \(P(6 \leq x \leq 14)\). How does this compare with the fraction in the interval \(\mu \pm 2 \sigma\) for any distribution? For mound-shaped distributions?

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