Respuesta :
Answer:
About 786 would be expected to yield more than 180 bushels of corn per acre
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this question, we have that:
[tex]\mu = 189.3, \sigma = 23.5[/tex]
Proportion of acres with more than 180 bushels of corn per acre:
This is 1 subtracted by the pvalue of Z when X = 180. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{180 - 189.3}{23.5}[/tex]
[tex]Z = -0.4[/tex]
[tex]Z = -0.4[/tex] has a pvalue of 0.3446.
1 - 0.3446 = 0.6554
Out of 1200:
0.6554*1200 = 786.48
About 786 would be expected to yield more than 180 bushels of corn per acre
