Answer to Question #149402 in Statistics and Probability for jee

Question

a. The manufacturer of aspirin claims that the proportion
of headache sufferers who get relief with
just two aspirins is 53%. What is the probability
that in a random sample of 400 headache sufferers,
less than 50% obtain relief? If 50% of the
sample actually obtained relief, what does this
suggest about the manufacturer’s claim?
b. Repeat part (a) using a sample of 1,000

Answer 1

From the claim we obtain that the probability to get a relief for a person is $0.53$.

a).The task is to find the probability that less than 200 sufferers will get relief. We will use the binomial distrubution with parameters 400 and $0.53$ (see https://mathworld.wolfram.com/BinomialDistribution.html). The probability that $k$ sufferers will get relief is: $p_k=P(X=k)=\frac{400!}{k!(400-k)!}(0.53)^k(0.47)^{400-k}$

$X$ denotes a random variable with the binomial distribution.

The aim is to find the probability: $\sum_{k=0}^{199}p_k=\sum_{k=0}^{199}\frac{400!}{k!(400-k)!}(0.53)^k(0.47)^{400-k}\approx$

$\approx0.1053$

The latter was rounded to 4 decimal places. We used the following command in Maple 13 to compute the sum:

sum(400!/(k!*(400-k)!)*(0.53)^k*(0.47)^(400-k),k=0..199);

Suppose that 50% obtained relief. In case we allow for the value to differ on $3\%$ from $53\%$ we can say that the claim is true. However, if we require the difference to be less than $3\%$, then the claim is false.

b). The task is to find the probability that less than $500$ sufferers will get relief. We use again the binomial distribution with parameters 1000 and $0.53$. The aim is to compute

$\sum_{k=0}^{499}\frac{1000!}{k!(1000-k)!}(0.53)^k(0.47)^{1000-k}\approx0.0267$

We computed it via the following command in Maple 13:

sum(1000!/(k!*(1000-k)!)*(0.53)^k*(0.47)^(1000-k),k=0..499);

The latter is rounded to 4 decimal places.

Answer: a). 0.1053; b). 0.0267