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STatiana [176]
3 years ago
7

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p? C

hoose the correct forumla below.
A. Upper E (Upper X )equalsStartRoot np (1 minus p )EndRoot

B. Upper E (Upper X )equalsp Superscript n

C. Upper E (Upper X )equals (1 minus p )Superscript n(1 minus p )Superscript n

D. Upper E (Upper X )equalsnp
Mathematics
1 answer:
charle [14.2K]3 years ago
5 0

Answer:

(D)E[ X ] =np.

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success​ p,

f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq  x\leq n

E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}

Now,

x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)

Substituting,

E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}

Factoring out the n and one p from the above expression:

E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}

Representing k=x-1 in the above gives us:

E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}

This can then be written by the Binomial Formula as:

E[ X ] = (np) (p +(1 - p))^{n -1 }= np.

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