Question

The highest grade for Florida oranges is U.S. Fancy and is based principally on color. Suppose that Shannon owns a small organic orchard and that the proportion of oranges she grows which can be classified as U.S. Fancy is 0.40 . One morning, Shannon randomly picks 44 oranges. Estimate the probability that the number of oranges that will be classified as U.S. Fancy, ???? , is fewer than 15. Use a normal approximation to the binomial distribution with continuity correction to obtain the solution. Give your answer precise to at least four decimal places.P(X < 16) =

Answer 1

Answer:

There is a 21.19% probability that the number of oranges that will be classified as U.S. Fancy is fewer than 15.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The binomial probability distribution is the probability of x sucesses on n trials, with a probability of p.

It has an expected value of

$E(X) = np$

And a standard deviation of:

$\sqrt{Var(X)} = \sqrt{np(1-p)}$

The binomial distribution can be approximated to the normal with n that is sufficiently large, with $\mu = E(X), \sigma = \sqrt{Var(X)}$.

In this problem, we have that:

The proportion of oranges she grows which can be classified as U.S. Fancy is 0.40. One morning, Shannon randomly picks 44 oranges. This means that $n = 44, p = 0.40$.

Estimate the probability that the number of oranges that will be classified as U.S. Fancy is fewer than 15.

This is the pvalue of Z when $X = 15$.

We have that:

$\mu = E(X) = np = 44*0.4 = 17.6$

$\sigma = \sqrt{Var(X)} = \sqrt{44*0.4*0.6} = 3.25$

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{15 - 17.6}{3.25}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119.

This means that there is a 21.19% probability that the number of oranges that will be classified as U.S. Fancy is fewer than 15.