Question

As reported in Trends in Television, the proportion of US households who have at least one VCR is 0.535. If 14 households are selected at random, without replacement, from all US households, what is the (approximate) probability that the number having at least one VCR is no more than 8 but at least 6.00. Be sure to use many decimal places in your calculations (at least 4), but report your answer to three decimal places.

Answer 1

Using the binomial distribution, it is found that there is a 0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

For each household, there are only two possible outcomes. Either it has at least one VCR, or it does not. The probability of a household having at least one VCR is independent of any other household, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this problem:

• 14 households, hence $n = 14$.
• 0.535 probability of having at least one VCR, hence $p = 0.535$.

The probability of at least 6 and no more than 8 is:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{14,6}.(0.535)^{6}.(0.465)^{8} = 0.1539$

$P(X = 7) = C_{14,7}.(0.535)^{7}.(0.465)^{7} = 0.2024$

$P(X = 8) = C_{14,8}.(0.535)^{8}.(0.465)^{6} = 0.2038$

Then:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8) = 0.1539 + 0.2024 + 0.2038 = 0.5601$

0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

A similar problem is given at https://brainly.com/question/24863377