Question

Cat behavior: A report stated that the average number of times a cat returns to it's good bowl during the day is 37. Assuming the variable is normally distributed with a standard deviation of 3, what is the probability that a cat would return to it's dish between 34 and 39 times a day. Use the graphing calculator. Round the answer to four decimal places.
P (34<X<39) =?​

Answer 1

Answer:

P (34<X<39) = 0.5899

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 37, \sigma = 3$

What is the probability that a cat would return to it's dish between 34 and 39 times a day.

This is the pvalue of Z when X = 39 subtracted by the pvalue of Z when X = 34. So

X = 39

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{39 - 37}{3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486

X = 34

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{34 - 37}{3}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.7486 - 0.1587 = 0.5899

So

P (34<X<39) = 0.5899