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 and standard deviation , the zscore of a measure X is given by:
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
And a standard deviation of:
The binomial distribution can be approximated to the normal with n that is sufficiently large, with .
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 .
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 .
We have that:
So
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.