Respuesta :
Answer:
0.3182 = 32.81% probability that a randomly selected acre has between 32647 and 43559 ants.
Step-by-step explanation:
When the distribution is normal, we use 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 = 42000, \sigma = 12275[/tex]
Find the probability that a randomly selectd acre has between 32647 and 43559 ants.
This is the pvalue of Z when X = 43559 subtracted by the pvalue of Z when X = 32647. So
X = 43559:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{43559 - 42000}{12275}[/tex]
[tex]Z = 0.13[/tex]
[tex]Z = 0.13[/tex] has a pvalue of 0.5517.
X = 32647:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32647 - 42000}{12275}[/tex]
[tex]Z = -0.76[/tex]
[tex]Z = -0.76[/tex] has a pvalue of 0.2236
0.5517 - 0.2236 = 0.3281
0.3182 = 32.81% probability that a randomly selected acre has between 32647 and 43559 ants.
