Question

In March the US unemployment rate was 4.4%. So, 4.4% of the US workforce (those ready, willing, and able to work) are unable to find a job. This does not include retired persons, disabled persons, or those that choose not to work. A random sample of 30 members of the workforce is obtained and the number that are unemployed is recorded.

(a) What is the distribution for the number of unemployed persons from 30 members of the workforce? State its two parameters.

(b) Find the probability that exactly 3 of the 30 selected are unemployed. Do not use statistical features of your calculator.

(c) Find the probability that between 4 and 10, inclusively, of the 30 selected are unemployed. You may use any features of your calculator.

(d) Compute the mean and standard deviation of the number unemployed persons from a sample of 30.

Answer 1

a)

this is binomial distribution

parameters are : n = 30 , p = 0.044

b)

Here, n = 30, p = 0.044, (1 - p) = 0.956 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 4)
P(X = 4) = 30C4 * 0.044^4 * 0.956^26
P(X = 4) = 0.0319

c)

Here, n = 30, p = 0.044, (1 - p) = 0.956, x1 = 4 and x2 = 10.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(4 <= X <= 10)
P(4 <= X <= 10) = (30C4 * 0.044^4 * 0.956^26) + (30C5 * 0.044^5 * 0.956^25) + (30C6 * 0.044^6 * 0.956^24) + (30C7 * 0.044^7 * 0.956^23) + (30C8 * 0.044^8 * 0.956^22) + (30C9 * 0.044^9 * 0.956^21) + (30C10 * 0.044^10 * 0.956^20)
P(4 <= X <= 10) = 0.0319 + 0.0076 + 0.0015 + 0.0002 + 0 + 0 + 0
P(4 <= X <= 10) = 0.0412

d)

mean = np
= 30 * 0.044 = 1.32

std.dev = sqrt(npq)
=sqrt(30 * 0.044 *(1-0.044))
= 1.1234