Question

William discovers that the number of nuts in a bag of cashews is normally distributed with mean 42 and standard deviation 4. If he buys 200 bags, how many of them can he expect to contain fewer than 34 nuts?

Answer 1

Answer:

4.56, so he should expect between 4 and 5 of them to contain fewer than 34 nuts.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean 42 and standard deviation 4.

This means that $\mu = 42, \sigma = 4$

Proportion with fewer than 34 nuts:

p-value of Z when X = 34. So

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

$Z = \frac{34 - 42}{4}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228.

Out of 200 bags:

0.0228*200 = 4.56.

4.56, so he should expect between 4 and 5 of them to contain fewer than 34 nuts.