Your answer would be 24/25
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.
In this question, plug in 14 for r and 8 for s, which will look like 3/98 + 5/64. Solve, and you will get 0.109 = 0.11.
Answer:
a) probability that is cracked=1/30 (3.33%)
b) probability that is discoloured = 29/600 (4.83%)
c) probability that is cracked and discoloured = 11/600 (1.83%)
Step-by-step explanation:
assuming that each stone is equally likely to be chosen then defining the events C= the stone is cracked , D= the stone is discoloured , N= the stone is neither cracked or discoloured, then
P(C)= number of favourable outcomes/total number of outcomes = 20 stones/600 stones = 1/30 (3.33%)
P(D)= number of favourable outcomes/total number of outcomes = 29 stones/600 stones = 29/600 (4.83%)
the probability that is discoloured and cracked is P(C∩D) , where
P(C∪D)=P(C) + P(D)-P(C∩D)
and
P(C∪D)= 1- P(N)
thus
1- P(N)=P(C) + P(D)-P(C∩D)
P(C∩D)= P(N)+P(C)+P(D) -1
replacing values
P(C∩D)= P(N)+P(C)+P(D)=562/600 + 20/600 + 29/600 -1= 611/600 -1 = 11/600
thus
P(C∩D)= 11/600 (1.83%)
Natural, Whole, and a rational number
