Math Problem Statement

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Part 1

In a certain​ area,

3737​%

of people own a dog. Complete parts a and b below.

Question content area bottom

Part 1

a. Find the probability that exactly

66

out of

1010

randomly selected people in the area own a dog.

The probability that exactly

66

out of

1010

randomly selected people in the area own a dog is

0.0850.085.

​(Type an integer or decimal rounded to three decimal places as​ needed.)

Part 2

b. In a random sample of

1010

people from this​ area, find the probability that

66

or fewer own a dog.

The probability that

66

or fewer out of

1010

randomly selected people in the area own a dog is

enter your response here.

​(Type an integer or decimal rounded to three decimal places as​ needed.)

Solution

Part 1:

We are given that 37% (or 0.37) of people in a certain area own a dog. The problem asks us to find the probability that exactly 6 out of 10 randomly selected people own a dog.

This is a binomial probability problem, where the probability of success (owning a dog) is p=0.37p = 0.37, the number of trials is n=10n = 10, and the number of successes (people owning a dog) is x=6x = 6.

The binomial probability formula is:

P(X=x)=(nx)px(1p)nxP(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}

Where:

  • (nx)\binom{n}{x} is the binomial coefficient (combination), which is calculated as n!x!(nx)!\frac{n!}{x!(n - x)!},
  • p=0.37p = 0.37 is the probability of owning a dog,
  • n=10n = 10 is the number of people,
  • x=6x = 6 is the number of successes (people owning a dog).

Let's compute the binomial probability:

P(X=6)=(106)(0.37)6(0.63)4P(X = 6) = \binom{10}{6} (0.37)^6 (0.63)^4

Using the formula for the binomial coefficient:

(106)=10!6!(106)!=10×9×8×74×3×2×1=210\binom{10}{6} = \frac{10!}{6!(10 - 6)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

Now calculate:

P(X=6)=210×(0.37)6×(0.63)4P(X = 6) = 210 \times (0.37)^6 \times (0.63)^4

Using a calculator:

P(X=6)0.085P(X = 6) \approx 0.085

So, the probability that exactly 6 out of 10 people own a dog is approximately 0.085.

Part 2:

Now, we need to find the probability that 66 or fewer out of 10 randomly selected people own a dog. This is the cumulative probability for X6X \leq 6, meaning we need to sum the probabilities of X=0,1,2,,6X = 0, 1, 2, \dots, 6.

This can be calculated using the binomial cumulative distribution function (CDF):

P(X6)=P(X=0)+P(X=1)++P(X=6)P(X \leq 6) = P(X = 0) + P(X = 1) + \dots + P(X = 6)

The cumulative probability can be calculated using statistical software or a binomial calculator. Using this method (or a standard calculator with a CDF function), we find:

P(X6)0.914P(X \leq 6) \approx 0.914

So, the probability that 66 or fewer out of 10 randomly selected people own a dog is approximately 0.914.

Final Answer:

  • Part 1: P(X=6)0.085P(X = 6) \approx 0.085
  • Part 2: P(X6)0.914P(X \leq 6) \approx 0.914

Would you like further clarification or details on the steps involved?

Here are 5 related questions:

  1. What is the binomial distribution formula and when is it used?
  2. How do you calculate cumulative probabilities for a binomial distribution?
  3. What are the differences between binomial and normal distributions?
  4. How do you find the mean and variance for a binomial distribution?
  5. Can the binomial distribution be approximated using a normal distribution?

Tip: When dealing with binomial probabilities, you can use a binomial calculator or statistical software (like Excel or a graphing calculator) to save time on computing the individual terms.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Probability Theory

Formulas

Binomial Probability Formula P(X = x) = C(n, x) * p^x * (1 - p)^(n - x)
Binomial Cumulative Distribution Function P(X <= x) = P(X = 0) + P(X = 1) + ... + P(X = x)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 9-11