Question

The childhood obesity rate in a certain Michigan city is 15 per- cent. Suppose that a random sample of 100 is drawn from the city for a research study. Let pˆ denote the proportion of obese children in the sample.

1. The standard deviation of pˆ is: --> A

(a) 0.036 (b) 0.227 (c) 0.773 (d) 0.933 (e) none

2. What is the probability that the sample proportion is at most .20 ( P (p ≤ 0.20)) ? --> D

(a) 0.081 (b) 0.227 (c) 0.773 (d) 0.919 (e) none

3. The probability that the sample proportion is within plus/minus 0.01 of the population proportion?

(a) 0.22 (b) 0.59 (c) 0.42 (d) 0.58 (e) 0.78

I have solved #1 and #2, but I am having trouble solving 3.

Answer 1

1) My work:
\[sd = \sqrt{\frac{ p (1-p) }{ n }} \] which is: \[= \sqrt{\frac{ (.15)(1-.15) }{ 100 }}\] = .0357

2) = normalcdf (a,b, mean, sd)
=normalcdf (-9^9, .20, .15, .0357)
= .9175