Problem 1 (S points total) The Department of Animal Regulation released informat

Question

Problem 1 (S points total) The Department of Animal Regulation released information on pet ownership for the population of all households in a particular county. The variable considered was (k) the number cf licensed dogs or cats for a household. The probability distribution for number of licensed dogs or cats for a household is given in the following table: x-#of licensed cats or dogs. 5 0.01 P(X x) 0.22 0.13 0.09 0.03 0.52 a. (2 points) Is this a legitimate probability model? Explain b. (3 points) What is the probability of randomly selecting a household with three or more licensed dogs or cats? Show work. Problem 2: (5 points) A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean 3 cm and standard deviation 0.1 cm. The specifications call for corks with diameters between 2.9 and 3.1 cm. A cork not meeting the specifications is considered defective. What is the probability of random selecting a defective cork? Show work.

Explanation / Answer

Problem 1

Part a

The probability model is legitimate if the sum of all probabilities is 1.

Here, P(X=x) = 0.52+0.22+0.13+0.09+0.03+0.01 = 1.00

So, this is a legitimate probability model.

Part b

We have to find P(X3)

P(X3) = P(X=3) + P(X=4) + P(X=5)

P(X3) = 0.09 + 0.03 + 0.01 = 0.13

Problem 2

We are given that diameters are normally distributed.

Mean = 3

SD = 0.1

We have to find the probability of randomly selecting a defective cork.

Specification limits are 2.9 and 3.1 cm.

We have to find P(X<2.9) + P(X>3.1)

First we have to find P(X<2.9)

Z = (X – Mean) / SD

Z = (2.9 – 3)/0.1 = -1

P(Z<-1) = P(X<2.9) = 0.158655254

Now, we have to find P(X>3.1)

P(X>3.1) = 1 – P(X<3.1)

Z = (3.1 – 3) / 0.1 = 1

P(Z<1) = P(X<3.1) = 0.841344746

P(X>3.1) = 1 – P(X<3.1)

P(X>3.1) = 1 – 0.841344746

P(X>3.1) = 0.158655254

P(X<2.9) + P(X>3.1) = 0.158655254 + 0.158655254 = 0.317310508

Required probability = 0.317310508

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