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4 years ago

7

Sarah Meeham blends coffee for Tasti-Delight. She needs to prepare 160 pounds of blended coffee beans selling for $4.23 per po

und. She plans to do this by blending together a high-quality bean costing $5.25 per pound and a cheaper bean at $2.00 per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean she should blend.

1 answer:

MissTica4 years ago

7 0

5$ because hes hella wet if you naam sayin

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nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

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

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

This Zscore is how many standard deviations the value of the measure X 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.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

So

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

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$

The standard deviation of a sample of n young man is given by the following formula

$s = \frac{\sigma}{\sqrt{n}}$

We want to have $s = 0.167$

$0.167 = \frac{2.8}{\sqrt{n}}$

$0.167\sqrt{n} = 2.8$

$\sqrt{n} = \frac{2.8}{0.167}$

$\sqrt{n} = 16.77$

$\sqrt{n}^{2} = 16.77^{2}$

$n = 281.23$

We need a sample of at least 282 young men.

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