This problem follows a binomial probability distribution because there are only two possible outcomes for each consumer: they either say it is important (success) or they do not (failure). The given parameters are:

• $n = 18$ (total consumers surveyed)
• $p = 0.45$ (probability of success, i.e., a consumer says it is important)
• $q = 1 - p = 0.55$ (probability of failure)

The binomial probability formula is:

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

where:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Now, we calculate each part.

(a) Probability of exactly 7 consumers saying it is important:

$P(X = 7) = \binom{18}{7} (0.45)^7 (0.55)^{11}$

(b) Probability of at least 10 consumers saying it is important:

$P(X \geq 10) = 1 - P(X \leq 9)$ This requires summing probabilities from $X = 0$ to $X = 9$ and subtracting from 1.

(c) Probability of fewer than 12 consumers saying it is important:

$P(X < 12) = P(X \leq 11)$ This requires summing probabilities from $X = 0$ to $X = 11$.

I will now compute these probabilities. scipy.stats import binom

Given data

n = 18 # Number of trials (consumers surveyed) p = 0.45 # Probability of success

(a) P(X = 7)

prob_7 = binom.pmf(7, n, p)

(b) P(X >= 10) = 1 - P(X <= 9)

prob_at_least_10 = 1 - binom.cdf(9, n, p)

(c) P(X < 12) = P(X <= 11)

prob_less_than_12 = binom.cdf(11, n, p)

Round to three decimal places

prob_7, prob_at_least_10, prob_less_than_12Here are the computed probabilities:

(a) $P(X = 7) = 0.166$
(b) $P(X \geq 10) = 0.253$
(c) $P(X < 12) = 0.946$

Would you like a deeper explanation or help with a similar problem?

Related Questions:

1. How do we derive the binomial probability formula?
2. What are the key assumptions behind using a binomial distribution?
3. How can we approximate binomial probabilities using a normal distribution?
4. What is the expected number of consumers who say it is important?
5. How would the probabilities change if the sample size increased to 30?

Tip: When dealing with large sample sizes, using the normal approximation to the binomial distribution can save time in calculations!