Question

In a study, 44% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Find the probability that when 14 adults are randomly selected, more than 3 are in excellent health.

Round to nearest ten-thousandth.  

Answer 1

Let , X be the number of adults that their health was excellent.

Here , X has binomial distribution with parameter n=14 and p=0.44

The pmf of X is ,

; x=0,1,2,......,n and q=1-p

Tn-1 P(X = x) = (%) pq"

= 0 ; otherwise

Now , the probability distribution table is ,

X	n	$p^x$	$q^{n-x}$	P(X=x)
0	1	1	0.000298	0.0003
1	14	0.44	0.000533	0.0033
2	91	0.1936	0.000951	0.0168
3	364	0.08518	0.001699	0.0527
4	1001	0.03748	0.003033	0.1138
5	2002	0.01649	0.005416	0.1788
6	3003	0.00726	0.009672	0.2108
7	3432	0.00319	0.017271	0.1892
8	3003	0.0014	0.030841	0.1301
9	2002	0.00062	0.055073	0.0682
10	1001	0.00027	0.098345	0.0268
11	364	0.00012	0.175616	0.0076
12	91	5.3E-05	0.3136	0.0015
13	14	2.3E-05	0.56	0.0002
14	1	1E-05	1	1E-05

Now , $P(X>3)=1-P(X\leq 3)$

$=1-[P(X=3)+P(X=2)+P(X=1)+P(X=0)]$

=1-(0.0527+0.0168+0.0033+0.0003)

=0.9269

Therefore , the probability that when 14 adults are randomly selected, more than 3 are in excellent health is 0.9269