Question

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Plz help

1. Identify proper model

2. Write prob mass function equations

Part 1. (8 points) For each random variable, state whether the random variable should be modeled with a Binomial distribution or a Poisson distribution. Explain your reasoning. State the parameter values that describe the distribution and give the probability mass function. Random Variable 1. A quality measurement for cabinet manufacturers is whether a drawer slides open and shut easily. Historically, 2% of drawers fail the easy slide test. A manufacturer samples 10 drawers from a batch. Assuming the chance of failure is independent between drawers, what type of distribution could be used to model the number of failed drawers from the sample of 10? Random Variable 2. The warranty for a particular system on a new car is 2 years. During which there is no limit to the number of warranty claims per car. Historically, the average number of claims per car during the period is 0.8 claims. What type of distribution could be used to model the number of warranty claims per car?

Answer 1

Random variable 1 -

The random variable should be modeled with a Binomial distribution because, it has the below properties.

• The experiment consists of 10 repeated trials.
• Each trial can result in just two possible outcomes - Drawers fail or do not fail.
• The probability of drawer failure, denoted by P = 0.02, is the same on every trial.
• The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

The parameter values of the distribution is n = 10. P = 0.02

The probability mass function is,

P(X = x) = ⁿC_x * 0.02^x * (1 - 0.02)^10-x = ⁿC_x * 0.02^x * 0.98^10-x

Random variable 2 -

The random variable should be modeled with a Poisson distribution because, it has the below properties.

• The experiment results in outcomes that can be classified as successes (warranty claims) or failures (or no warranty claims).
• The average number of successes ($\lambda$) that occurs in a specified period is known. Thus, $\lambda$ = 0.8
• The probability that a success (warranty claims) will occur is proportional to the size of the period.
• The probability that a success (warranty claims) will occur in an extremely small period is virtually zero.

The parameter of the distribution is $\lambda$ = 0.8.

The probability mass function is,

P(X = x) = (e^-0.8) (0.8^x) / x!