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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

Your friend has a bag of 20 marbles. Seven

Elis [28]

Answer:

20%

Step-by-step explanation:

First, I added them up. From the things we know, blue, red, green, yellow combined equals 16 marbles. 20-16 equals 4. So there are 4 white marbles. 4/20 times 5 equals 20/100. So the percent of the marbles that are white is 20%.

6 0

3 years ago

Read 2 more answers

Please help me out . :)

Veseljchak [2.6K]

Answer:

Only 1 pair of feet duhh and a few of the other things that are not in the same way as the same thing is true for the natives from the americas and now claim it as if it were ares to start with the same thing as the last time I was there and I was just wondering if you had any ideas on how to get a new one for the natives from the americas and now claim it as if it were ares to start with the same thing as the last time I was in there middle of school all as but it 4th 4images in a group and that will not have make of these days available until the next time week have the case or the natives time Europe and fend it will take place in a actuality for my new place job then the same time you

5 0

3 years ago

Read 2 more answers

What is the equation of a line with a slope of 2 and a y-intercept of -5?

RSB [31]

Y=2x-5

Hope this helps ya :D

7 0

2 years ago

3 questions in 1

MrRissso [65]

Answer:

1. D. 20, 30, and 50

2. A. 86

3. B. 94

Step-by-step explanation:

1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.

The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.

Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.

Thus:

Q1 = (85 + 87)/2 = 86

Q3 = (93 + 95)/2 = 94

IQR = Q3 - Q1 = 94 - 86

IQR = 8

Outliers in the data set are data values below the lower limit or above the upper limit.

Let's find the lower and upper limit.

Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74

The data values below the lower limit (74) are 20, 30, and 50

Let's see if we have any data value above the upper limit.

Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106

No data value is above 106.

Therefore, the only outliers of the data set are:

D. 20, 30, and 50

2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.

Thus:

Q1 = (85 + 87)/2 = 86

3. Q3 = (93 + 95)/2 = 94

3 0

2 years ago