Question

A tree in an orchard has 200 oranges. Of the oranges, 40 are not ripe. Use the normal distribution to approximate the binomial distribution, and determine the probability a box containing 35 oranges has at most two oranges that are not ripe.

What is the probability that the box containing 35 oranges has at most two unripe oranges?
a) 0.11
b) 0.24
c) 0.36
d) 0.49

Answer 1

To determine the probability that a box containing 35 oranges has at most two oranges that are not ripe, we can use the normal approximation. We can calculate the mean and standard deviation of the binomial distribution and then find the z-score for a box with 35 oranges. By looking up the probability corresponding to the z-score in the normal distribution table, we can determine the final probability.

Step-by-step explanation:

To approximate the binomial distribution, we can use the normal distribution. Given that there are 200 oranges in the tree and 40 are not ripe, we can say that the ripe oranges follow a binomial distribution with parameters n = 200 and p = 160/200. To determine the probability that a box containing 35 oranges has at most two oranges that are not ripe, we can use the normal approximation. The mean of the binomial distribution is given by μ = np and the standard deviation is given by σ = sqrt(np(1-p)). We can use these values to calculate the z-score for a box with 35 oranges and then look up the corresponding probability in the normal distribution table.

1. Calculate the mean of the binomial distribution: μ = np = 200 * (160/200) = 160.
2. Calculate the standard deviation of the binomial distribution: σ = sqrt(np(1-p)) = sqrt(200 * (160/200) * (40/200)) = sqrt(64) = 8.
3. Calculate the z-score for a box with 35 oranges: z = (x - μ) / σ = (35 - 160) / 8 ≈ -18.75.
4. Look up the probability corresponding to the z-score -18.75 in the normal distribution table.