The distribution that could be used to model the number of vacuum cleaners that break down during the year is the binomial distribution. The relevant parameters for the binomial distribution are the number of trials (n), which is the total number of vacuum cleaners sold (17), and the probability of success (p), which is the chance of a vacuum cleaner breaking down (7% or 0.07).
To calculate the probability that more than 2 vacuum cleaners will break down, we need to sum the probabilities of 3 or more successes in the binomial distribution.
Using the formula for the probability mass function (PMF) of the binomial distribution:
P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
where X is the random variable representing the number of vacuum cleaners that break down.
Calculating each term separately:
P(X = 0) = (17 choose 0) * (0.07^0) * (0.93^17) ≈ 0.155
P(X = 1) = (17 choose 1) * (0.07^1) * (0.93^16) ≈ 0.342
P(X = 2) = (17 choose 2) * (0.07^2) * (0.93^15) ≈ 0.339
Summing these probabilities:
P(X > 2) = 1 - (0.155 + 0.342 + 0.339) ≈ 0.164
Therefore, the probability that more than 2 vacuum cleaners will break down during the year is approximately 0.164 or 16.4%.
As for using a normal approximation, the binomial distribution can be approximated by a normal distribution when both np and n(1-p) are greater than or equal to 5. In this case, np = 17 * 0.07 = 1.19 and n(1-p) = 17 * 0.93 = 15.81. Since both values are less than 5, it would not be reasonable to use a normal approximation for this calculation.