Respuesta :
Answer:
a. 0.443
b. 0.023
Step-by-step explanation:
Problems of normal distributions 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 weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds.
This means that [tex]\mu = 22, \sigma = 5[/tex]
a. Find the probability that a randomly selected turkey weighs between 20 and 26 pounds.
This is the pvalue of Z when X = 26 subtracted by the pvalue of Z when X = 20. So
X = 26
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{26 - 22}{5}[/tex]
[tex]Z = 0.8[/tex]
[tex]Z = 0.8[/tex] has a pvalue of 0.788
X = 20
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 22}{5}[/tex]
[tex]Z = -0.4[/tex]
[tex]Z = -0.4[/tex] has a pvalue of 0.345
0.788 - 0.345 = 0.443
The answer is 0.443
b. Find the probability that a randomly selected turkey weighs below 12 pounds.
This is the pvalue of Z when X = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 22}{5}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.023
The answer is 0.023.
