The ages of alligators in a swamp are normally distributed, with a mean of 50 years and a standard deviation of 4. Approximately what percentage of the alligators are between 45 and 55 years old?

78.87%

10.56%

89.44%

67.76%

The percentage is 78.88%

So, Option A is correct.

Step-by-step explanation:

Mean = 50

Standard Deviation = 4

Percentage of the alligators are between 45 and 55 years old = ?

We need to find z-score when x = 45 and when x = 55

For x = 45

z-score = x - Mean/Standard Deviation

z-score = 45-50/4

z-score = - 1.25

P (z=-1.25) can be found by using z-score table

P (z = - 1.25) = 0.1056

For x = 55

z-score = x - Mean/Standard Deviation

z-score = 55-50/4

z-score = 1.25

P (z = 1.25) can be found by using z-score table

P (z = 1.25) = 0.8944

So, area between 45 and 55 is area between z = - 1.25 and z = 1.25

Subtracting the values of P (z = - 1.25) from P (z = 1.25)

0.8944 - 0.1056 = 0.7888

So, the percentage is 78.88%

So, Option A is correct.

A

Step-by-step explanation:

i just took it and made a 90