Betsy has 10 more dimes than quarters if she has 3.45 how many coins does she have

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Find f(-2) for f(x) = 2(4)^x

stepladder [879]

Answer:

2(4)^-2

=2/4^2

2/16

1/8

thanks

3 0

2 years ago

Read 2 more answers

A pair of boots has an original price of $95. they are on sale for 35 percent off. what number can you multiply $95 by to equal

Alex Ar [27]

Multiply it by .35
.35 is equal to %35.

4 0

3 years ago

How would I solve this equation for mass (m)?

zaharov [31]

Simple...

you have: $F= \frac{G m^{2} }{ d^{2} }$

You want to find m-->>>

1.)Isolate m

$d^{2} *F= \frac{G m^{2} }{ d^{2} } * d^{2}$

Leaving you with...

F* $d^{2}$ = $G m^{2}$

Keep isolating m-->>>

$F* d^{2} =G m^{2}$

$\frac{F* d^{2} }{G} = \frac{G m^{2} }{G}$

Square root to solve what just m is-->>>

$\sqrt{ \frac{F* d^{2} }{G} } = \sqrt{ m^{2} }$

= $\sqrt{ \frac{F* d^{2} }{G} } =m$

That is how you solve for m...

Thus, your answer.

7 0

2 years ago

(3x-2)(3x-2) solve the problem using factoring

Effectus [21]

9x squared - 6x - 6x + 4
9x squared - 12x + 4

5 0

2 years ago