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guapka [62]
3 years ago
9

West and Company uses the LIFO method to calculate inventory values. Their beginning inventory consisted of 200 units at a cost

of $9.00 each. Purchases include 300 units at $10.00 each on February 18; 400 units at $11.00 each on July 16; and 100 units at $12.00 each on December 5. If there were 300 units remaining in imventory at the end of the year, what was the cost of goods sold?
Mathematics
1 answer:
nirvana33 [79]3 years ago
7 0
The total inventory is 200 + 300 + 400 + 100 = 1000. If 300 units remained, the we calculate the cost of the 700 sold via LIFO. These 700 units include the 100 units at $12.00, the 400 units at $11.00, and 200 units at $10.00 (out of the 300 purchased). This is a total cost of 100*12 + 400*11 + 200*10 = 1200 + 4400 + 2000 = $7600.
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0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 can be 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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