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Anna71 [15]
3 years ago
15

A survey claims that the percent of an entire population that agrees with redeveloping the city park is likely between 49.1% and

57.5%. The remainder of the people in the survey were against redevelopment. How many people were surveyed?
Mathematics
1 answer:
kumpel [21]3 years ago
7 0

Answer:

<u>567</u> people were surveyed.

Step-by-step explanation:

Given:

A survey claims that the percent of an entire population that agrees with redeveloping the city park is likely between 49.1% and 57.5%.

Now, to find total people who were surveyed.

<em>According to survey percent of population agrees between = 49.1% to 57.5%.</em>

So, we need to find the error of margin:

\frac{57.5\%-49.1\%}{2}

=\frac{8.4\%}{2}

=4.2\%.

Let the number of people surveyed be n.

Now, to get the number of people surveyed:

4.2\%=\frac{1}{\sqrt{n} }

\frac{4.2}{100}=\frac{1}{\sqrt{n} }

0.042=\frac{1}{\sqrt{n} }

<em>By cross multiplying we get:</em>

<em />\sqrt{n}=\frac{1}{0.042}<em />

<em />\sqrt{n}=23.81<em />

<em>Using square root on both sides we get:</em>

n=566.91.

<u>Approximately:</u>

<u />n=567.<u />

Therefore, 567 people were surveyed.

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Lina20 [59]

Answer:

(a) The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b) The most likely value for <em>X</em> is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number among the four homeowners  who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.

The random sample of homeowners selected is, <em>n</em> = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.

The probability mass function of <em>X</em> is:

P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.33\\=1.32

Thus, the most likely value for <em>X</em> is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

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Step-by-step explanation:

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