Question

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A survey conducted for the Northwestern National Life Insurance Company revealed that 70% of American workers say job stress caused frequent health problems. Suppose a random sample of 10 American workers is selected.

What is the probability that more than seven of them say job stress caused frequent health problems?

What is the probability that exactly five say job stress caused frequent health problems?

What is the expected [mean] number of workers that would say job stress caused frequent health problems?

[Hint: Use binomial distribution to answer question]

Answer 1

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 10 * 0.7
= 7
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 10 * 0.7 * 0.3
= 2.1
III.
standard deviation = sqrt( variance ) = sqrt(2.1)
=1.4491

a.
the probability that more than seven of them say job stress caused frequent health problems
P( X < = 7) = P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 7 ) * 0.7^7 * ( 1- 0.7 ) ^3 + ( 10 6 ) * 0.7^6 * ( 1- 0.7 ) ^4 + ( 10 5 ) * 0.7^5 * ( 1- 0.7 ) ^5 + ( 10 4 ) * 0.7^4 * ( 1- 0.7 ) ^6 + ( 10 3 ) * 0.7^3 * ( 1- 0.7 ) ^7 + ( 10 2 ) * 0.7^2 * ( 1- 0.7 ) ^8 + ( 10 1 ) * 0.7^1 * ( 1- 0.7 ) ^9 + ( 10 0 ) * 0.7^0 * ( 1- 0.7 ) ^10   
= 0.6172
P( X > 7) = 1 - P ( X <=7) = 1 -0.6172 = 0.3828
b.
the probability that exactly five say job stress caused frequent health problems
P( X = 5 ) = ( 10 5 ) * ( 0.7^5) * ( 1 - 0.7 )^5
= 0.1029
c.
the expected [mean] number of workers that would say job stress caused frequent health problems = 10*0.7 =7