MATH SOLVE

2 months ago

Q:
# 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) =

Accepted Solution

A:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]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 [tex]E(X) = np[/tex]And a standard deviation of:[tex]\sqrt{Var(X)} = \sqrt{np(1-p)}[/tex]The binomial distribution can be approximated to the normal with n that is sufficiently large, with [tex]\mu = E(X), \sigma = \sqrt{Var(X)}[/tex].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 [tex]n = 44, p = 0.40[/tex].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 [tex]X = 15[/tex].We have that:[tex]\mu = E(X) = np = 44*0.4 = 17.6[/tex][tex]\sigma = \sqrt{Var(X)} = \sqrt{44*0.4*0.6} = 3.25[/tex]So[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{15 - 17.6}{3.25}[/tex][tex]Z = -0.8[/tex][tex]Z = -0.8[/tex] 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.