Question

More than a decade ago, high levels of lead in the blood put 81% of children at risk. A concerted effort was made to remove lead from the enviornment. Now suppose only 17% of children in the United States are at risk of high blood- level leads.

A. In a random sample of 186 children taken more than a decade ago, what is the probability that 50 or more had high blood levels?

B. In a random sample of 186 children taken now, what is the probability that 50 or more have high blood levels?

Answer 1

A)

P = 0.81

N = 186

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 150.66

N*(1-p) = 35.34

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/(s.d)

Mean = n*p = 150.66

S.d = √{n*p*(1-p)} = 5.35027102117

P(x>=50)

By continuity correction

P(x>49.5)

Z = (49.5 - 150.66)/5.35027102117

Z = -18.9

From z table, P(z>-18.9) = 1

B)

Here mean = n*p = 180*0.17 = 30.6

S.d = √{n*p*(1-p)} = 5.03964284448

P(x>49.5)

Z = (49.5 - 30.6)/5.03964284448

Z = 3.75

From z table, P(z>3.75) = 0.000088 ~ 0.0001