It is known that 15% of the population is afraid of being alone at night. If a random sample of 20 Americans is selected, what is the probability that :
a. Exactly 6 of them are afraid?
b. Less than 2 of them are afraid?
c. At least 1 of them is afraid?
d. What is the standard deviation for the number of people that are afraid?
a)
Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 6
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X = 6)
P(X = 6) = 20C6 * 0.15^6 * 0.85^14
P(X = 6) = 0.0454
b)
Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X <= 2).
P(X <2) = (20C0 * 0.15^0 * 0.85^20) + (20C1 * 0.15^1 *
0.85^19)
P(X <2) = 0.0388 + 0.1368
P(X < 2) = 0.1756
c)
Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X >= 1).
We need to calculate P(X <= 0).
P(X <= 0) = (20C0 * 0.15^0 * 0.85^20)
P(X <= 0) = 0.0388
P(X <= 0) = 0.0388
P(X>=1) = 1- P(x< =0)
=1 - 0.0388
= 0.9612
d)
std.dev = sqrt(npq)
=sqrt(20 * 0.15 * 0.85)
= 1.5969
c)
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