Question

The spending at Target is distributed normally with a mean spending of $47.67 and a standard deviation of $5.50. What is the probability that the spending is between 46 and 49.56 dollars?

Answer 1

Answer:

25.10% probability that the spending is between 46 and 49.56 dollars

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 47.67, \sigma = 5.5$

What is the probability that the spending is between 46 and 49.56 dollars?

This is the pvalue of Z when X = 49.56 subtracted by the pvalue of Z when X = 46. So

X = 49.56

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

$Z = \frac{49.56 - 47.67}{5.5}$

$Z = 0.34$

$Z = 0.34$ has a pvalue of 0.6331

X = 46

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

$Z = \frac{46 - 47.67}{5.5}$

$Z = -0.3$

$Z = -0.3$ has a pvalue of 0.3821

0.6331 - 0.3821 = 0.2510

25.10% probability that the spending is between 46 and 49.56 dollars