Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought

This question is poorly phrased

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.

The mean provided is: 12.5

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.

μ - σ

12.5 - 2.4

= 10.1

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

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

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)

= 12.5 - 7.2

= 5.3

The calculations indicate that

5.3 years is at one side of 99.7%

Therefore,

100 - 99.7%/2 = 0.3%/2

= 0.15%

Moreover, 10.1 years corresponds to one side of 68%

Thus

100 - 68%/2 = 32%/2 = 16%

Consequently, the percentage of a lion living between 5.3 to 10.1 years is evaluated as 16% - 0.15%

= 15.85%

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

is represented as a decimal =

= 15.85/ 100

= 0.1585

Response: Using the Pythagorean theorem, we can set up the equation as 14^2=9^2+x^2

                                     -9^2    -9^2