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Harrizon [31]
3 years ago
12

Consider the experiment of tossing a fair coin three times and observing the number of heads that result (X = number of heads).

What is the standard deviation for this distribution?
Mathematics
1 answer:
Crank3 years ago
7 0

Answer:

The standard deviation for this distribution is 0.8660.

Step-by-step explanation:

For each coin toss, there are only two possible outcomes. Either it is heads, or it is tails. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The standard deviation of the binomial distribution is:

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

In this problem, we have that:

In each toss, heads or tails are equally as likely, since the coin is fair. So p = \frac{1}{2} = 0.5

Three throws, so n = 3

What is the standard deviation for this distribution?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{3*0.5*0.5} = 0.8660

The standard deviation for this distribution is 0.8660.

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divide both sides by 3

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1. Janessa is playing a board game with two friends. Using a single die, one friend rolled a four, and the other friend rolled a
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1)    3, 4

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Based on aâ poll, among adults who regret gettingâ tattoos, 18â% say that they were too young when they got their tattoos. Assum
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Answer:

a) 20.44% probability that none of the selected adults say that they were too young to get tattoos.

b) 35.90% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

c) 56.34% probability that the number of selected adults saying they were too young is 0 or 1.

d) No

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they say they were too young when they got their tattoos, or they don't say that. Each adult is independent of each other. So we use the binomial probability distribution to solve this question.

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.

18% say that they were too young when they got their tattoos.

This means that p = 0.18

Eight adults who regret getting tattoos are randomly selected

This means that n = 8

a. Find the probability that none of the selected adults say that they were too young to get tattoos.

This is P(X = 0).

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

P(X = 0) = C_{8,0}.(0.18)^{0}.(0.82)^{8} = 0.2044

20.44% probability that none of the selected adults say that they were too young to get tattoos.

b. Find the probability that exactly one of the selected adults says that he or she was too young to get tattoos.

This is P(X = 1).

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

P(X = 1) = C_{8,1}.(0.18)^{1}.(0.82)^{7} = 0.3590

35.90% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

c. Find the probability that the number of selected adults saying they were too young is 0 or 1.

Either a. or b.

20.44 + 35.90 = 56.34

56.34% probability that the number of selected adults saying they were too young is 0 or 1.

d. It we randomly select 9 adults. Is 1 a significantly low number who day that they were too young to get tattoos?

Now n = 9

It is significantly low if it is more than 2.5 standard deviations below the mean.

The mean is E(X) = np = 9*0.18 = 1.62

The standard deviation is \sqrt{V(X)} = \sqrt{n*p*(1-p)} = \sqrt{9*0.18*0.82} = 1.15

1 > (1.62 - 2.5*1.15)

So the answer is no.

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<h3></h3><h3>How the sale can manager estimate the number of math books </h3>

Since extra math's  books will be needed after giving out the available math's books.

Hence:

Additional maths books=10%× 253,625

Additional maths books=25,362 books

Based on the above calculation the manager should carryout survey based on the number of people per class and provide them with 25,362 books.

Therefore how can the sale manager estimate the number of math books to be sure that there are more then enough is that the manager should carryout survey based on the number of people per class and provide 25,362 books.

Learn more about How the sale can manager estimate the number of math books here:brainly.com/question/18558957

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