Question

The loaves of rye bread distributed to a local store by a certain bakery have an average length of 30 centimeters and a standard deviation of 2 centimeters. Assuming the lengths are normally distributed. What percentage of loaves are between 26.94 and 32.18 centimeters

Answer 1

Answer:

79.91% of loaves are between 26.94 and 32.18 centimeters

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:

$\mu = 30, \sigma = 2$

What percentage of loaves are between 26.94 and 32.18 centimeters

This is the pvalue of Z when X = 32.18 subtracted by the pvalue of Z when X = 26.94.

X = 32.18:

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

$Z = \frac{32.18 - 30}{2}$

$Z = 1.09$

$Z = 1.09$ has a pvalue of 0.8621

X = 26.94:

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

$Z = \frac{26.94 - 30}{2}$

$Z = -1.53$

$Z = -1.53$ has a pvalue of 0.0630

0.8621 - 0.0630 = 0.7991

79.91% of loaves are between 26.94 and 32.18 centimeters