Need help to solve question

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

True or False? Natural Numbers are closed under division.

nignag [31]

False because you can subtract your way out of the set of whole numbers

7 0

3 years ago

An airline charges an extra fee if a suitcase weighs more than 50 pounds. After packing, Li's suitcase weighs 47.75 pounds. Whic

Vinvika [58]

Answer:

47.75 + x Less-than-or-equal-to 50

= 47.75 + x ≤ 50

Step-by-step explanation:

Solving the above Question:

Not going over the 50 pound case mean, less than or equal to 50 pounds

Let the extra pound of weight be represented as x

Hence, the inequality equation that can be used to determine how much more weight can be added to the suitcase without going over the 50-pound weight limit =

47.75 + x ≤ 50

4 0

2 years ago

Read 2 more answers

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali

Dominik [7]

Answer:

y 0 1 2 3

P(Y=y) 0.0676 0.3549 0.3875 0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

Y(W,W) Y(W,T) Y(W,F) Y(W,S)

Y(T,W) Y(T,T) Y(T,F) Y(T,S)

Y(F,W) Y(F,T) Y(F,F) Y(F,S)

Y(S,W) Y(S,T) Y(S,F) Y(S,S)

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

= 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

= (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

= 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

=(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

= 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)

= (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

= 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y 0 1 2 3

P(Y=y) 0.0676 0.3549 0.3875 0.19

The PMF plot is attached as a photo here.

7 0

3 years ago

Write an equation and solve.

mezya [45]

Equation: 50+15x=200
solution: x= 10

it would take him 10 weeks to purchase the bike he wants.

8 0

3 years ago