Answer: 0.1312
Step-by-step explanation:
Given: The proportion of population has a retirement account : p = 0.48
Sample size : n = 632
Let q be th sample proportion.
The probability that the proportion of persons with a retirement account will differ from the population proportion by greater than 3% will be :-
[tex]P(|q-p|>0.03)=1-P((|q-p|\leq 0.03)\\\\=1-P(-0.03<q-p<0.03)\\\\=1-P(\dfrac{-0.03}{\sqrt{\dfrac{(0.48)(1-0.48)}{632}}}<\dfrac{q-p}{\sqrt{\dfrac{p(1-p)}{n}}}<\dfrac{0.03}{\sqrt{\dfrac{(0.48)(1-0.48)}{632}}})\\\\=1-P(-1.5096<z<1.5095) \ \ \ \ [\ Z=\dfrac{q-p}{\sqrt{\dfrac{p(1-p)}{n}}}\ ]\\\\=1-(2P(Z<1.5095)-1)\ \ \ \ [P(-z<Z<z)=2(Z<z)-1]\\\\=2-2P(Z<1.5095)=2-2( 0.9344)\ \ \ [\text{by p-value table}]\\\\=0.1312[/tex]
Hence, the required probability = 0.1312
