the answer would be 8.
2 times 8 = 16 minus 5 = 11
Answer:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic 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 all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z =
~ N(0,1)
where,
= average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(
12)
P(
12) = P(
) = P(Z
1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Answer:
The total cost of the companies will be the same after 6 visits. The cost will be $40 after 6 visits
Step-by-step explanation:
Answer:
The predicted GPA is then y = 0.149(15) + 0.89 = 3.125
Step-by-step explanation:
Although you don't specifically say so, the equation you provide here is probably a "best fit" equation based upon data: GPA versus number of hours of study per week.
Here, y = 0.149x + 0.89 and the number of study hours of interest is 15.
The predicted GPA is then y = 0.149(15) + 0.89 = 3.125