A woman who is 64 inches tall has a shoulder width of 16 inches. How tall is a woman who has a shoulder width of 18.5 inches?

1 answer:

7 0

If we turn this to a ratio we get 64:16 and we need ?:18.5. To easily solve this we would divide 64 by 16 and 64/16= 4 now that we know that the shoulder width is 1/4 if a persons height we can then multiply 18.5 by 4 to find the woman’s height. 18.5x4=74
So your answer is 74 inches

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A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Qua

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Answer:

There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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.

In this problem, we have that:

The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that $\mu = 120, \sigma = 0.05$ .

Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that $n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005$ .