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ValentinkaMS [17]
3 years ago
13

John is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than u

sual. When he weighs one of the sticks, he finds that it is 2.33 oz. The manufacturer's website states that the average weight of each stick is 1.75 oz with a standard deviation of 0.22 oz. Assume that the weight of the drumsticks is normally distributed. What is the probability of the stick's weight being 2.33 oz or greater? Give your answer as a percentage precise to at least two decimal places. You might find this table of standard normal critical values useful.
Business
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

0.0042 is the probability of the stick's weight being 2.33 oz or greater.  

Explanation:

We are given the following information in the question:

Mean, μ = 1.75 oz

Standard Deviation, σ = 0.22 oz

We are given that the distribution of drumsticks is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(stick's weight being 2.33 oz or greater)

P(x > 2.33)

P( x > 2.33) = P( z > \displaystyle\frac{2.33 - 1.75}{0.22}) = P(z > 2.6363)

= 1 - P(z \leq 2.6363)

Calculation the value from standard normal z table, we have,  

P(x > 2.33) = 1 - 0.9958 =0.0042= 0.42\%

0.0042 is the probability of the stick's weight being 2.33 oz or greater.

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Answer:

The answer is:

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Total share            <u>$39, 500</u>                            <u>$50, 500</u>

Interest        (10% * $80, 000)                           (10% * $40, 000)

Profit share (50% * $43,000)                           (50% * $43,000)

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