A public health organization reports that 40%of baby boys 6-8 months old in the United
States weigh more than 20 pounds. A sample of 10 babies is studied. Round the answers to three decimal places.
what is the probability that more than 4 weigh more than 20 pounds
What is the probability that fewer than 3 weigh more than 20 pounds?
Would it be unusual if more than 7 of them weigh more than 20 pounds?

Answers

Answer 1

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

What is the binomial distribution formula?

The formula is:

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

x is the number of successes.n is the number of trials.p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

[tex]P(X > 4) = 1 - P(X \leq 4)[/tex]

In which:

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

Then:

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

[tex]P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061[/tex]

[tex]P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403[/tex]

[tex]P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209[/tex]

[tex]P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150[/tex]

[tex]P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502[/tex]

Hence:

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325[/tex]

[tex]P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675[/tex]

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

[tex]P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)[/tex]

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

[tex]P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106[/tex]

[tex]P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016[/tex]

[tex]P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001[/tex]

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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The first column is labeled sum with entries 5, 7, 9, 11, 13, 15, 17.

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We have given,

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The sum is the forword counting of the numbers.

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The second column is labeled frequency with entries 1, 2, 3, 4, 3, 2, 1.

The follow me to get 50 points.

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