.125/100 converted to a decimal

Why are there three partial products in 481x352 and only two partial products in 481x302?

Sindrei [870]

352 can be separated by 300, 50, and 2. These numbers can produce 3 partial products.

302 can only produce 2 partial products because the only numbers to consider are 300 and 2. 0 is not considered because any number multiplied by 0 is equal to 0.
481 ( 400 + 80 + 1)
<u> x352 </u> (300 + 50 + 2)
2 (2 x 1)
160 (2 x 80)
800 (2 x 400) Partial product 1 = 962
50 (50 x 1)
4000 (50 x 80)
20000 (50 x 400) Partial product 2 = 24,050
300 (300 x 1)
24000 (300 x 80)
<u> 120000 (</u>300 x 400) Partial product 3 = 144,300
169,312

3 0

3 years ago

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.9 ounces and stand

fgiga [73]

Answer:

91.60% probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 7.9, \sigma = 0.16, n = 3, s = \frac{0.16}{\sqrt{3}} = 0.0924$

What is the probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces?

This is the pvalue of Z when X = 8.09 subtracted by the pvalue of Z when X = 7.76. So

X = 8.09

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

By the Central Limit Theorem

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