Respuesta :
Answer:
(a) 0.7472
(b) 0.2397
(c) 3.6
Step-by-step explanation:
Let X denote the number of children who are hyperlipidemic.
The proportion of children who are hyperlipidemic is, p = 0.30.
A random sample of n = 12 children are analyzed.
Every child is independent of the others to be expected to meet the criteria for hyperlipidemia.
The random variable X follows a binomial distribution with parameters n = 12 and p = 0.30.
(a)
Compute the probability that at least 3 are hyperlipidemic as follows:
[tex]P(X\geq 3)=1-P(X<3)\\\\=1-\sum\limits^{2}_{0}{{12\choose x}(0.30)^{x}(0.70)^{12-x}}\\\\=1-0.25282\\\\=0.74718\\\\\approx 0.7472[/tex]
Thus, the probability that at least 3 are hyperlipidemic is 0.7472.
(b)
Compute the probability that exactly 3 are hyperlipidemic as follows:
[tex]P(X=3)={12\choose 3}(0.30)^{3}(0.70)^{12-3}\\\\=220\times 0.027\times 0.040353607\\\\=0.23970042558\\\\\approx 0.2397[/tex]
Thus, the probability that exactly 3 are hyperlipidemic is 0.2397.
(c)
Compute the expected number of children who would meet the criteria for hyperlipidemia as follows:
[tex]E(X)=np=12\times 0.30=3.6[/tex]
Thus, 3.6 children would be expected to meet the criteria for hyperlipidemia.
