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3 years ago

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A city is running out of phone numbers available for their population. They currently have 2,000,000 available phone numbers. Th

ey intend to create new area codes with the form ABC, where A can be digits 2-8, B can be digits 5-9, and C can be digits 5-9. How many TOTAL phone numbers will the city have after adding the new area codes?

1 answer:

PIT_PIT [208]3 years ago

4 0

Answer:

The total phone numbers, if the latter ABC is followed, will be 175.

Step-by-step explanation:

We are told that the digits will be in the form ABC and A can take values from 2 – 8, B from 5 – 9 and C from 5 – 9 as well.

This means:

A has 7 possible values (when you count the number of numbers between 2 and 8).

B has 5 possible values.

C has 5 possible values.

Therefore, the total phone numbers, if the latter ABC is followed, will be:

7 x 5 x 5 = 175.

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According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For

SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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:

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

The Z-score measures how many standard deviations the measure 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.71$

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here $n = 16$ , we want $P(X \geq 13)$ . So

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591$

$P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835$

$P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273$

$P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042$

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741$

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now $n = 160$ . So

$\mu = E(X) = np = 160*0.71 = 113.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74$

Using continuity correction, this is $P(X \geq 110 - 0.5) = P(X \geq 109.5)$ , which is 1 subtracted by the pvalue of Z when X = 109.5. So

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

$Z = \frac{109.5 - 113.6}{5.74}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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