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Alla [95]
3 years ago
15

A firm that stocks lightbulbs gathers the following information: Demand = 19,500 units per year, Ordering cost = $25 per order,

Holding cost = $4 per unit per year. The firm wants to calculate the: a) EOQ for the lightbulbs. b) Annual holding costs for the lightbulbs. c) Annual ordering costs for the lightbulbs.
Mathematics
1 answer:
vredina [299]3 years ago
3 0

Answer:

Step-by-step explanation:

Demand = 19,500 units per year (D)

Ordering cost = $25 per order (O)

Holding cost = $4 per unit per year (C)

a) EOQ=\sqrt{ \frac{2\times D\times O}{C}}

   EOQ=\sqrt{ \frac{2\times 19,500\times 25}{4}}

   = \sqrt{\frac{975000}{4}}

   = \sqrt{243,750}

   = 493.71044 ≈ 494

b) Annual holding cost = 4\times(\frac{Q}{2})

                                      = 4\times(\frac{494}{2})

                                      = 4 × 247

                                      = 988

c) Annual ordering cost = O\times(\frac{D}{Q} )

                                       = 25\times(\frac{19,500}{494} )

                                       = 25 × 39.47

                                       = 986.75

AOQ = 494

Annual holding cost = 988

Annual ordering cost = 986.75

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To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data
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.

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

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\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

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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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