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Zigmanuir [339]
3 years ago
5

A parent volunteer group is raising money by making custom hats to sell at school activities they plan to sell the hats for $11

each hat costs $5 to make they spent $50 advertising
Mathematics
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

  • (11 - 5)n - 50

Step-by-step explanation:

  • Cost of 1 hat = $5
  • Selling price = $11
  • Advertising = $50

Let the number of hats be n. Then money made is

<u>Profit for each hat:</u>

  • (11-5)

<u>Profit for n hats </u>

  • (11 - 5)n

<u>Deduct cost of advertisement - 50</u>

  • (11 - 5)n - 50

Option C is correct

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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
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Answer:

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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 normal distributions can be solved using the z-score formula.

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

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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.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

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Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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Z = \frac{77 - 74}{1.2274}

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

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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