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1 month ago

Assume there are 365 days in a year.

Mathematics

1 answer:

babunello [11.8K]1 month ago

6 0

Response:

1) The likelihood that ten students in a classroom have distinct birthdays is 0.883.

2) The likelihood that at least two out of ten students share a birthday is 0.002.

Detailed explanation:

Given: Assuming 365 days are in a year.

To find: 1) What is the probability that ten students in a classroom have unique birthdays?

2) What is the probability that at least two among ten students have a birthday in common?

Solution:

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Total outcomes = 365

1) The chance that ten students in a class hold different birthdays is

The first student can have their birthday on any of the 365 days, the second can only on 364/365 and so forth...

$\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883$

The chance that ten students in a class have distinct birthdays is 0.883.

2) The likelihood that at least two out of ten students share a birthday

P(2 on the same day) = 1 - P(2 not on the same day)

$\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]$

$\text{P(2 born on same day) }=1-[\frac{364}{365}]$

$\text{P(2 born on same day) }=0.002$

The probability that at least two students in a class have the same birthday is 0.002.

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Complete Question

The lifespans of lions at a specific zoo follow a normal distribution. The average lifespan is 12.5 years with a standard deviation of 2.4 years. Apply the empirical rule (68-95-99.7%) to estimate the likelihood of a lion living between 5.3 and 10.1 years.

Answer:

The likelihood of a lion living from 5.3 to 10.1 years is 0.1585

Step-by-step explanation:

According to the empirical rule:

1) 68% of the data falls within 1 standard deviation of the mean, meaning between μ - σ and μ + σ.

2) 95% of the data is contained within 2 standard deviations around the mean - between μ - 2σ and μ + 2σ.

3) 99.7% of the data lies within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

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Standard deviation: 2.4 years

Starting with the first rule:

1) 68% falls within 1 standard deviation from the mean, implying between μ - σ and μ + σ.

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We now apply the second rule:

2) 95% of the data lies within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

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Now applying the last rule:

3)99.7% of the data resides within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

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The calculations indicate that

5.3 years is at one side of 99.7%

Therefore,

100 - 99.7%/2 = 0.3%/2

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Thus, the estimated probability of a lion surviving between 5.3 and 10.1 years

is represented as a decimal =

= 15.85/ 100

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