Question

A survey of 3,219 consumers by GfK NOP for the Food Standards Agency showed that 49% of respondents correctly identified the use by date as the best indicator of whether food is safe to eat or not. Suppose a random sample of 1,100 consumers is taken and each is asked a number of questions on food safety at home. What is the probability that more than 550 consumers identify the use by date as the best indicator of whether food is safe to eat or not?

Answer 1

n = 1100

p = 0.49

x = 550

So,

P(x > 550) = 1100C550 * 0.49^550 * 0.51^550 +........+ 1100C1100 * 0.49^1100 * 0.51^0

P = 0.244

Answer 2

We can model the number of consumers who correctly identify the use by date as a binomial random variable with parameters $n = 1100$ and $p = 0.49$, where $n$ is the sample size and $p$ is the proportion of consumers in the population who correctly identify the use by date.

Let $X$ be the number of consumers in the sample who correctly identify the use by date. Then, $X$ follows a binomial distribution with parameters $n = 1100$ and $p = 0.49$, i.e., $X \sim \text{Bin}(1100, 0.49)$.

We want to find $P(X > 550)$, which is the probability that more than 550 consumers in the sample correctly identify the use by date.

We can use the normal approximation to the binomial distribution to approximate $X$ by a normal distribution with mean $\mu = np = 1100 \times 0.49 = 539$ and standard deviation $\sigma = \sqrt{np(1-p)} \approx 12.67$.

Then, we have:

P(X>550)≈P(Z>σ550−μ)=P(Z>12.67550−539)≈P(Z>0.87)≈0.194

$\mathrm{<span\; style="font-family:\; sans-serif;">where\; \$Z\$\; is\; a\; standard\; normal\; random\; variable.</span>}$

Therefore, the probability that more than 550 consumers in the sample correctly identify the use by date as the best indicator of whether food is safe to eat or not is approximately 0.194.