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In a high school there are 1200 students. Estimate the probability that more than 130 students were born in January under each of the following assumptions. You do not have to use the continuity correction.

(a) Months are equally likely to contain birthdays
(b) Days are equally likely to be birthdays

Respuesta :

Answer:

The probability that more than 130 students were born in January under the given assumptions is 0.00086

Step-by-step explanation:

Let X be a variable denoting the number of birthdays in January.

P(that a gien birthday is in the month of January)    =   [tex]\frac{1}{12}[/tex]

                                                                 E(X) = np = [tex]1200\times\frac{1}{12}[/tex] = 100                                                                                                                                                                                          

                                                                   σ = [tex]\sqrt{ (np ( 1 -p ))}[/tex] = 9.5743

Now, as per question

P (X> 130) = 1 - P( x ≤ 130)

                 = 1 - P (Z ≤  [tex]\frac{130 - 100}{9.53742}[/tex] )

                  = 1 - 0.99914

                  =0.00086

MrRoyal

The probability that more than 130 students were born in January, if

  • Months equally likely contain birthdays is 0.087%
  • Days equally likely contain birthdays is 0.18%

(a) Months equally likely contain birthdays

There are 12 months in a year.

So, the probability of having a birthday in a January is

[tex]p = \frac 1{12}[/tex]

The number of students (n) is given as 1200.

So, the mean is:

[tex]\mu =np[/tex]

[tex]\mu =1200 \times \frac 1{12}[/tex]

[tex]\mu =100[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{np(1 - p)}[/tex]

[tex]\sigma = \sqrt{1200 \times \frac 1{12} \times (1 - \frac 1{12})}[/tex]

[tex]\sigma = 9.574[/tex]

Calculate the z-score for x = 130

[tex]z = \frac{x - \mu}{\sigma}[/tex]

[tex]z = \frac{130 - 100}{9.574}[/tex]

[tex]z = \frac{30}{9.574}[/tex]

[tex]z = 3.133[/tex]

So, the probability is represented as:

[tex]P(x > 130) = P(z > 3.133)[/tex]

From z-scores of probabilities, we have:

[tex]P(x > 130) = 0.00086515[/tex]

Express as percentage

[tex]P(x > 130) = 0.086515\%[/tex]

Approximate

[tex]P(x > 130) = 0.087\%[/tex]

Hence, the probability is 0.087%

(b) Days equally likely contain birthdays

There are 365 days in a year, ad 31 days in January.

So, the probability of having a birthday in January is

[tex]p = \frac {31}{365}[/tex]

[tex]p = 0.0849[/tex]

The number of students (n) is given as 1200.

So, the mean is:

[tex]\mu =np[/tex]

[tex]\mu =1200 \times 0.0849[/tex]

[tex]\mu =101.88[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{np(1 - p)}[/tex]

[tex]\sigma = \sqrt{1200 \times 0.0849 \times (1 - 0.0849)}[/tex]

[tex]\sigma = 9.656[/tex]

Calculate the z-score for x = 130

[tex]z = \frac{x - \mu}{\sigma}[/tex]

[tex]z = \frac{130 - 101.88}{9.656}[/tex]

[tex]z = \frac{28.12}{9.656}[/tex]

[tex]z = 2.912[/tex]

So, the probability is represented as:

[tex]P(x > 130) = P(z > 2.912)[/tex]

From z-scores of probabilities, we have:

[tex]P(x > 130) = 0.0017956[/tex]

Express as percentage

[tex]P(x > 130) = 0.17956\%[/tex]

Approximate

[tex]P(x > 130) = 0.18\%[/tex]

Hence, the probability is 0.18%

Read more about probabilities at:

https://brainly.com/question/13031543

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