# Find the mean and standard deviation for this binomial distribution? I need help calculating binomial random variable?If x is a binomial random variable, calculate the probability of x if N= 13,...

Find the mean and standard deviation for this binomial distribution?

I need help calculating binomial random variable?

If x is a binomial random variable, calculate the probability of x if N= 13, P=0.40 A. What is the probability that x is at least 3? B. What is the probability that x is no more than 10? C. Find the mean and standard deviation for this binomial distribution?

Asked on by mlunar74

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `n=13,p=.4`

** For a binomial experiment, `p(x=a)=""_nC_ap^aq^(n-a)` ; we assume the requiements for a binomial experiment have been met.

(1) Find the probability that x is at least 3

`P(x>=3)=1-[P(x=0)+P(x=1)+P(x=2)]`

So `P(x>=3)=""1-[_13C_0*.4^0*.6^13+_13C_1*.4^1*.6^12+_13C_2*.4^2*.6^11]`

`=1-[(1)(1)(.00131)+(13)(.4)(.00218)+(78)(.16)(.00363)]`

`~~1-[.05790]~~.9421`

*`_nC_r` is sometimes written `([n],[r])` , and the formula is `(n!)/((n-r)!r!)`

** Note that we used the complement -- this makes the computations easier. We could have directly computed the probability for each x from 4 to 13 and gotten the sum -- yikes!

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So the probability that x is at least 3 is approximately .9421

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(2) Probability that x is no more than 10 or `P(x<=10)` :

`P(x<=10)=1-[P(x=11)+P(x=12)+P(x=13)]`

`=1-[+_13C_11*.4^11*.6^2+_13C_12*.4^12*.6^1+_13C_13*.4^13*.6^0]`

`=1-[(78)(.00004)(.36)+(13)(.00002)(.6)+(1)(.000006)(1)]`

`=1-.00129~~.9987`

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The probability that x is no more than 10 is approximately .9987

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(3) The mean `mu` is found by `mu-np` so `mu=13*.4=5.2`

**Note that you do not have to memorize this formula. If I told you that 40% of the group have a quality, and that there were 13 members of the group, you would say that 5.5 have that quality

The standard deviation is found by `sigma=sqrt(npq)` where q is the complement of p. Thus `sigma=sqrt(13(.4)(.6))~~1.77`

*** You could approximate the answers to (1) and (2) using the normal approximation to the binomial distribution.

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