Print Full Name: Recitation Day and Time: Week 6-Pre-Recitation Homework Chapter

Question

Print Full Name: Recitation Day and Time: Week 6-Pre-Recitation Homework Chapter 12 Random Variables- 19 points total Print out this document and write your answers directly on this page OR you may write your answers on notebook paper. This is due at the start of your recitation Wednesday November 1" or T November 2d. Because I at 3 PM either to me in class or under my door in WNGR 215. am getting this up late, if you need more time, you can turn in by November 3" Problem 1: (5 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 (x) the number of 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. 0.52 0.22 0.13 0.09 0.03 0.01 (2 points) Is this a legitimate probability model? Explain a. (3 points) What is the probability of randomly selecting a household with three or more licensed dogs or cats? Show work b. 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 randomly selecting a defective cork? Show work

Explanation / Answer

Mean of diameter = 3 cm
Standard deviation of diameter = 0.1 cm
The cork is not defective if the diameter lie between 2.9 cm and 3.1 cm

Probability that cork is not defective = P(2.9 < X < 3.1)

Z value for X = 2.9 cm is given as,

Z = (2.9 - 3) / 0.1 = -1

Z value for X = 3.1 cm is given as,

Z = (3.1 - 3) / 0.1 = 1

So, probability that cork is not defective = P(2.9 < X < 3.1)

= P( - 1 < Z < 1) = P(Z < 1) - P(Z < -1)

= 0.8413 - 0.1587

= 0.6826

Probability that cork is not defective = 0.6826

Probability of a randomly selected defective cork = 1 - 0.6826 = 0.3174

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