It is believed that 58% of Americans like to eat at restaura

Pam Stokes

Pam Stokes

Answered question

2021-12-14

It is believed that 58% of Americans like to eat at restaurants every week. You interview 6 Americans and ask if they like to eat at restaurants every week. The random variable x represents the number of Americans that like to eat at restaurants every week.
Find the mean of the binomial distribution. μ=?
Find the standard deviation of the binomial distribution. σ=?
What is the probability that at least 1 of the Americans surveyed likes to eat at restaurants every week? Provide an answer to 4 decimal places.

Answer & Explanation

Melissa Moore

Melissa Moore

Beginner2021-12-15Added 32 answers

Step 1
Given that:
Number of observations =n=6
Probability of success =p=58%=0.58
So, X follows binomial distribution with n=6 and p=0.58
mean=μ=np=6(0.58)=3.48
standard deviation
=σ
=np(1p)
=6(0.58)(10.58)
=1.4616
=1.20896650078
1.2090
Step 2
At least 1 means 1 or more.
We will use the complement property of probability for this
P(X1)
=1P(X=0)
=1(6C0)(0.58)0(10.58)6
=10.005489031744
=0.994510968256
0.9945
Answer:
a) 3.48
b) 1.2090
c) 0.9945
John Koga

John Koga

Beginner2021-12-16Added 33 answers

Step 1
Sample size n=6
Let x is a binomial variate that denotes the no of Americans that like to cat at restaurants every week
Success probability
p=58%=0.58
q=1p=0.42
By using Binomal distribution
P(x=r)=nCrprqnr
=6Cr(0.58)r(0.42)6r
a) Mean =np
m=6(0.58)=3.48
b) Standard derivation σ=npq=6(0.58)(0.42)
=1.2089
c) P (atleasti) =P(x1)
=1p(x<1)
=1p(x=0)
=16C0(0.58)0(0.42)60
=10.0055
=0.9945

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