A veterinary researcher takes an SRS of 60 horses presenting with colic whose average age is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of horses coming to the veterinary clinic is 8 years. The probability that a sample mean is 12 or larger for a sample from the horse population is:

the probability of that the sample mean will have 12year or larger is 0.4013,.

Step-by-step explanation:

The probability that a sample mean is 12 or larger for a sample from the horse population is:

Given,

N, number of sample = 60 horses

x₁, variate mean = 12 years.

x, mean age = 10 years.

σ, standard deviation = 8 years.

To calculate for the probability, we are going to use the formula of normal distribution.

normal distribution is calculated by:

z=[ x₁-x] / σ

where z is called the normal standard variate,

x₁ is the value of the variable,

x is the mean value of the distribution and

σ is the standard deviation of the distribution.

The z-value corresponding to 12year is

z=[ x₁-x] / σ

z=[12 - 10] / 8 = 2/8=0.25

we will use the table of normal distribution to find the corresponding z value.

From Table of z, the area corresponding to a z-value of 0.25 is 0.0987.

The total area under the standardized normal curve is unity and since the curve is symmetrical,

The area to the right of the z=0.25 ordinate is 0.5000-0.0987 = 0.4013.

Thus, the probability of the sample mean is 0.4013, for 60 horses, it is likely that 60 * 0.4013 = 24.08 = 24 horses will have 12year or larger.