Question

Let X, denote the weight of a randomly selected prepackaged one-pound bag of carrots. Of course, one pound bags of carrots won't weigh exactly one pound. In fact, history suggests that Xi is normally distributed with a mean of 1.18 pounds and a standard deviation of 0.07 pound. Now, let W denote the weight of randomly selected prepackaged three-pound bag of carrots. Three pound bags of carrots won't weigh exactly three pounds either. In fact, history suggests that Wis normally distributed with a mean of 3.22 pounds and a standard deviation of 0.09 pound. Selecting bags at random, what is the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag?

Answer 1

Answer:

0.9830 = 98.30% probability that the sum of three one-pound bags exceeds the weight of one three-pound bag

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.

Three one pound bags:

3 one pound mags, which have mean 1.18 and standard deviation of 0.07. The mean is 1.18 multiplied by 3, and the standard deviation will be 0.07 multiplied by the square root of 3. So

$\mu_{1} = 3*1.18 = 3.54$

$\sigma_{1} = 0.07\sqrt{3} = 0.1212$

Three pound bags:

Mean 3.22, standard deviation 0.09. So

$\mu_{2} = 3.22, \sigma_{2} = 0.09$

What is the probability that the sum of three one-pound bags exceeds the weight of one three-pound bag?

Here, we work with the subtraction of the normal variables. The mean is the subtraction of the means, and the standard deviation is the square root of the sum of the variances. So

$\mu = \mu_{1} - \mu_{2} = 3.54 - 3.22 = 0.32$

$\sigma = \sqrt{\sigma_{1}^2 + \sigma_{2}^2} = \sqrt{0.1212^2+0.09^2} = 0.151$

This is the probability of the subtraction being positive, that is, 1 subtracted by the pvalue of Z when X = 0. Then

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

$Z = \frac{0 - 0.32}{0.151}$

$Z = -2.12$

$Z = -2.12$ has a pvalue of 0.0170

1 - 0.0170 = 0.9830

0.9830 = 98.30% probability that the sum of three one-pound bags exceeds the weight of one three-pound bag