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Lubov Fominskaja [6]
3 years ago
14

According to a​ report, 67.5​% of murders are committed with a firearm. ​(a) If 200 murders are randomly​ selected, how many wou

ld we expect to be committed with a​ firearm? ​(b) Would it be unusual to observe 153 murders by firearm in a random sample of 200 ​murders? Why?
Mathematics
1 answer:
kolezko [41]3 years ago
8 0

Answer: The answer is 135 murders.

Step-by-step explanation: The report tells us that statistically 67.5% of murders are committed using a firearm. It follows therefore that in a sample of 200 randomly selected murders, one would expect that 67.5% of those would be by a firearm. \frac{67.5}{100} * 200 = 135.

It would certainly be higher that the expected value based on previous data collected but it would not be unusual because one sample may have a higher than "normal" amount of murders by firearm. Statistics aren't going to be exact for every sample.

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What percentage of babies born in the United States are classified as having a low birthweight (<2500g)? explain how you got
lawyer [7]

Answer:

2.28% of babies born in the United States having a low birth weight.

Step-by-step explanation:

<u>The complete question is</u>: In the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g. What percent of babies born in the United States are classified as having a low birth weight (< 2,500 g)? Explain how you got your answer.

We are given that in the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g.

Let X = <u><em>birth weights of newborn babies</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 3,500 g

            \sigma = standard deviation = 500 g

So, X ~ N(\mu=3500, \sigma^{2} = 500)

Now, the percent of babies born in the United States having a low birth weight is given by = P(X < 2500 mg)

         

   P(X < 2500 mg) = P( \frac{X-\mu}{\sigma} < \frac{2500-3500}{500} ) = P(Z < -2) = 1 - P(Z \leq 2)

                                                                 = 1 - 0.97725 = 0.02275 or 2.28%

The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.

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3 years ago
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Step-by-step explanation:

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3 years ago
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Gre4nikov [31]

Answer:

Step-by-step explanation:

Simplifying

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Reorder the terms:

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Reorder the terms:

8 + -3a = -12 + 2z

Solving

8 + -3a = -12 + 2z

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-8' to each side of the equation.

8 + -8 + -3a = -12 + -8 + 2z

Combine like terms: 8 + -8 = 0

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Simplifying

a = 6.666666667 + -0.6666666667z

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Answer:

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Step-by-step explanation:

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