# 34%  of women consider themselves fans of professional baseball. You randomly select six women and ask each if she considers herself a fan of professional baseball.   Complete parts (a) through  (c)(a) Find  the mean of the binomial distribution.     _______  ( Round to the nearest tenth as needed)(b) Find the variance of the binomial distribution.     ________  ( Round to the nearest tenth as needed)(c) Find the standard deviation of the binomial distribution.     ________  ( Round to the nearest tenth  as needed.)Can  you please break this math problem down for me so I can understand better. Thank you.

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34%  of women consider themselves fans of professional baseball. You randomly select six women and ask each if she considers herself a fan of professional baseball.   Complete parts (a) through  (c)

(a) Find  the mean of the binomial distribution.

_______  ( Round to the nearest tenth as needed)

(b) Find the variance of the binomial distribution.

________  ( Round to the nearest tenth as needed)

(c) Find the standard deviation of the binomial distribution.

________  ( Round to the nearest tenth  as needed.)

Can  you please break this math problem down for me so I can understand better. Thank you.

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Step 1

Introduction:

Let X denotes the number of women who consider themselves as fans of professional baseball, in the sample.

Now, when a woman is asked whether she is a fan or not, there may be two responses: “yes” and “no”. Consider the response yes as a “success” and the response no as a “failure”.

As 34% are self-proclaimed fans of professional baseball, the probability of “success” is 34% or 0.34. Denote this probability of success as p. So, p = 0.34.

The sample collected is of size 6, so that, n = 6. It is safe to assume that the response of one woman in the sample (regarding, whether she is a baseball fan or not) is independent of that of another. So, the probability of success is the same for all women sampled.

Now, if the possible outcomes of a trial can be classified as a “success” and a “failure” and the probability of success of each trial, p is a constant, then, for n of those trials, the number of successes, X has a Binomial distribution with parameters n and p. The mean or expectation of the Binomial random variable is np and the variance is np(1 – p).

Step 2

Parts a, b and c:

a:

The mean of the Binomial distribution is: np = 6 ∙ (0.34) = 2.04 ≈ 2.0 (to the nearest tenth).

b:

The variance of the Binomial distribution is: np(1 – p) = 6 ∙ (0.34) ∙ (1 – 0.34) = 6 ∙ (0.34) ∙ (0.66) = 1.3464 ≈ 1.3 (to the nearest tenth).

c:

The standard deviatio...

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