Question

According to a poll, 5% of adults travelled out of the state in 2015. A random sample of 60 adults is selected and let X be the number of adults that travelled out of the state in 2015. (a) Name the probability distribution for X. State the parameters. (b) Find the probability that less than 2 adults travelled out of the state in 2015. Do not evaluate your result. (c) Find the mean, the variance, and the standard deviation of X. (d) Name an appropriate discrete distribution to approximate the probability in (b). State the parameter(s) and compute the approximate probability. (e) Is the approximation in (d) justified? Explain.

Answer 1

(a) The name of the probability distribution is Binomial Distribution with parameter n = 60, p = 5% = 0.05

(b) Here we have to find $P(X<2)=P(X=0)+P(X=1)=\sum_{j=0}^{1} {60 \choose j}(0.05)^j (1-0.05)^{60-j}$

(c) Mean = $np = 60 \times 0.05 = 3$ , Variance = $np(1-p) = 60 \times 0.05 \times 0.95 = 2.85$ and SD is $\sqrt{ 2.85} = 1.688$

(d) The Poisson Distribution is to approximate the probability.

Because Poisson distribution is derived from Binomial Distribution approximately when n is too large and comparatively p is too small. Here in this case n=60, p=0.05 and hence justified. And the parameter of the distribution will be then $\lambda = np = 60 \times 0.05 = 3$

So we have to calculate $P(X<2)=P(X=0)+P(X=1)=\sum_{j=0}^{1} \dfrac{e^{-3}3^j}{j!} = e^{-3}(1+3) = 4e^{-3}$

N.B. An interesting property of Poisson Distribution is that its mean and variance are equal, here we have noticed that $2.85 \approx 3$