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*If an exercise asks you to use R, include a copy of the code and output. Please edit your code and output to be only the relevant portions.

*If a problem does not specify how to compute the answer, you many use any appropriate method. I may ask you to use R or use manually calculations on your exams, so practice accordingly.

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*Be sure to submit the HWK4 Autograde Quiz which will give you ~20 of your 40 accuracy points.

*50 points total: 40 points accuracy, and 10 points completion

Discrete RV Expectation, Variance, and Transformation

Exercise 1: A chemical supply company ships a certain solvent in 10-gallon drums. Let X represent the number of drums ordered by a randomly chosen customer. Assume X has the following probability mass function (pmf). The mean and variance of X is : \(\mu_X=2.2\) and \(\sigma^2_X=1.76=1.32665^2\):

X P(X=x)
1 0.4
2 0.3
3 0.1
4 0.1
5 0.1
  1. Calculate \(P(X \le 2)\) and describe what it means in the context of the problem. \(P(X \le 2) = 0.70\) This means that there is a 70% chance of a customer buying two or less drums of solvent.
  1. Let Y be the number of gallons ordered, so \(Y=10X\). Complete the probability mass function of Y.
y P(Y=y)
10 0.4
20 0.3
30 0.1
40 0.1
50 0.1
  1. Calculate the mean number of gallons ordered \(\mu_Y\).

\(\mu_Y = \mu_X * 10 = 22 Gallons\)

  1. Calculate the standard deviation of the number of gallons ordered, \(\sigma_Y\).

\(\sigma_Y = \sigma_X * 10 = 13.2665 Gallons\)

Normal RVs

Exercise 2: The bonding strength \(S\) of a drop of plastic glue from a particular manufacturer is thought to be well approximated by a normal distribution with mean 98 lbs and standard deviation 7.5 lbs. \(S~\sim N(98, 7.5^2)\). Compute the following values using a normal model assumption.

  1. What proportion of drops of plastic glue will have a bonding strength between 95 and 104 kg according to this model?

Assuming that there is a typo in the question, and it is supposed to be 95-104 lbs instead of kg: Approximately 44.35% of drops of plastic glue will have a bonding strength between 95 and 104 lbs, assuming a normal distribution. If this is not a typo, the proportion would be zero.

  1. A single drop of that glue had a bonding strength that is 0.5 standard deviations above the mean. What proportion of glue drops have a bonding strength that is higher?

Approximately 30.85% of glue drops have a bonding strength that is higher.

  1. What bonding strength did a drop of glue have that is at the 90th percentile?

A glue drop at the 90th percentile would have an approximate bonding strength of 107.6 lbs.

  1. What is the IQR of bonding strength for drops of glue from this manufacturer?

The IQR of bonding strength would be 103.025-92.975 = 10.05 lbs.

  1. Drops of a similar plastic glue from another manufacturer (manufacturer B) is claimed to have bonding strength well approximated by a normal distribution with mean 43 kg and standard deviation of 3.5 kg \(W_{B.kg}~\sim N(43, 3.5^2)\). Transform the bonding strength of manufacturer B into lbs using the conversion: 1 kg \(\approx\) 2.20462 lbs. You can use the transformation: \(W_{B.lbs}=2.20462*W_{B.kg}\). Compare the shape, center, and spread of the two glues’ bonding strength.

\(W_{B.lbs}~\sim N(94.79866, 7.71617^2)\) whereas \(W_{A.lbs}~\sim N(98, 7.5^2)\)

Sampling Distributions

Exercise 3: A serving of a specific type of yogurt has a sugar content that is well approximated by a Normally distributed random variable \(X\) with mean 13 g and variance: \(1.3^2 g^2\). We can consider each serving as an independent and identical draw from X.

  1. In what percent of servings will the sugar content be above 13.3 g?

The sugar content will be above 13.3 g in approximately 40.87% of servings.

  1. What is the probability that a randomly chosen serving will have a sugar content between 13.877 and 12.123? What do we call the difference: 13.877-12.123=1.754?

There is a 50% chance of randomly choosing a serving within this range. The range is identified as the IQR.

  1. Calculate the probability that in 6 servings, only 1 has a sugar content below 13 g.

There is a 9.375% chance of only 1 in 6 servings having a sugar content below 13 g.

  1. Describe the sampling distribution for the mean sugar content of 6 servings \(\bar{X}\). (Give shape, mean, and standard deviation or variance, if possible)

Shape: Symmetrical, Sample Mean: 13 g, Sample SD: 0.5307 g.

  1. What is the interquartile range of the sampling distribution for the sample mean \(\bar{X}\) when n=6? Is that value larger or smaller than the IQR implied in part b? Why does the relative sizes of the IQRs make sense?

IQR = 13.358 - 12.642 = 0.716. The IQR is smaller in this scenario compared to part b. This makes sense because the standard deviation is also smaller in this situation.

  1. What is the probability that the mean sugar content in 6 servings is more than 13.3 g ?

The probability of the mean in six servings being above 13.3 g is 0.2859.

  1. Is it more or less likely that the mean sugar content is above 13.3 g in 10 servings or 6 servings (as computed in f)? Can you explain it without actually computing the new probability?

It is less likely when sampling 10 servings instead of 6 because the standard deviation will decrease the more servings are accounted for.

  1. Suppose each large yogurt container of this type contains 10 servings and consider the total sugar content in each container as a sum of 10 iid random draws from \(X \sim N(13, 1.3^2)\). If you were to eat a whole large container of yogurt, above what total sugar content would you consume with 95% probability? Show and briefly explain your calculations.

Mean total sugar: \(T = 10*13 = 130 g\) Standard Deviation: sd(T) = \(sqrt(10*(1.3^2)) = 4.1110\) Finding normal probability with 5% chance: > qnorm(0.05,130,4.1110) = 123.238 g

Exercise 4: You will be comparing the sampling distributions for two different estimators of \(\sigma\), the population standard deviation.

When trying to estimate the standard deviation of a population (\(\sigma\)) from a sample we could use:

The graphs below give the sampling distributions produced by these estimators when drawing a sample of size 8 from a normal population with mean \(\mu_x=3\) and standard deviation \(\sigma_X=5\).

  1. What do you notice about the mean of the standard deviations produced using the \(s_1\) estimator compared to the \(s_2\) estimator compared to the true population standard deviation? Why do we prefer to use the \(s_1\) formulation when we have a sample of data and are interested in estimating the population standard deviation? (You should use the resulting histograms to help you answer the question and use the word “bias”.)

We would use the s1 sampling distribution method because there is less bias between the mean and actual values of standard deviation.