Let x be a binomial random variable with n=20 and p=42/100. Calculate the mean, variance and standard deviation of the random variable x.

emancipezN

emancipezN

Answered question

2021-08-04

Let x be a binomial random variable with n=20 and p=42100.
a) Calculate the mean, variance and standard deviation of the random variable x.
b) Use the results of part a to calculate the intervals μ±σ, μ±2σ, and μ±3σ. Find the probability that an observation will fall into each of these intervals.
c) Are the results of part b consistent with Tchebysheff’s Theorem? With the Empirical Rule? Why or why not?

Answer & Explanation

Ian Adams

Ian Adams

Skilled2021-08-17Added 163 answers

Step 1
Considering that, allow x to be a binomial random variable with n=20 and p=42100
The function of the binomial probability distribution is
P[X=x]=P(x)
P[X=x]=(ab)pxqnx; x=0, 1,  , n, 0<p<1, q=1p
=0. otherwise
The binomial distribution's mean is np.
E(X)=n× p
E(X)=20×42100
E(X)=8.4
The binomial distribution function's variance is npq.
Var(X)=δ2=npq
Var(X)=20×42100×(142100)
Var(X)=20×0.42×0.58
Var(X)=4.872
Standard deviation is (sd)=Var(X)
sd=δ=Var(X)
δ=4.872
δ=2.2073
Step 2
Calculate the intervals utilizing the results from section a. μ±σ, μ±2σ, and μ±3σ.
The confidence interval for μ±σ
C.I.=(μ=δ, μ+δ)
C.I.=(8.42.2073, 8.4+2.2073)
C.I.=(6.1927, 10.6073)
The Confidence interval for μ±2σ
C.I.=(μ2δ, μ+2δ)
C.I.=(8.42×2.2073, 8.4+2×2.2073)
C.I.=(3.9854, 12.8146)
The confidence interval for μ±3σ is
C.I.=(μ3δ, μ+3δ)
C.I.=(8.43×2.2073, 8.4+3×2.2073)
C.I.=(1.7781, 15.0219)
The likelihood that an observation will fall within each of these periods may be determined.

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