Respuesta :
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.
