Question

In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p?

(a)$n=10,p=0.5$

(b)$n=40,p=0.88$

(c)$n=100,p=0.2$

(d)$n=100,p=0.99$

(e)$n=1000,p=0.003$

Short answer

Because the required requirements are only met in option(c), option(c) is the correct response. $n=100,p=0.2$

Step-by-step solution

Given information 

Given in the question that, which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p

Explanation

The following conditions must be met in order to approximate a binomial distribution using a normal distribution:

$np\ge 10$

$nq\ge 10$

(a) $\mathrm{np}=10(0.5)=5$, $\mathrm{nq}=\mathrm{n}(1-\mathrm{p})=10(1-0.5)=5$

(b) localid="1649527995119" $\mathrm{np}=40(0.88)=35.2,\mathrm{nq}=\mathrm{n}(1-\mathrm{p})=40(1-0.88)=4.8$

(c) $\mathrm{np}=100(0.2)=20,\mathrm{nq}=\mathrm{n}(1-\mathrm{p})=100(1-0.2)=80$

(d) $np=100(0.99)=99,nq=n(1-p)=100(1-0.99)=1$

(e)$\mathrm{np}=1000(0.03)=3,\mathrm{nq}=\mathrm{n}(1-\mathrm{p})=1000(1-0.003)=997$