Midterm<\/p>\n
Morgan Holland<\/p>\n
5\/22\/2020<\/p>\n
Instructions<\/p>\n
You will need alcohol. RData dataset to complete the Midterm.<\/p>\n
The dataset includes data on 9,822 individuals. Each individual report various demographic and health characteristics, as well as a variable that equals 1 if the person reports abusing alcohol.<\/p>\n
Question 1.<\/p>\n
Load the alcohol.RData dataset. Convert it to a table named \u201calcohol.\u201d Use print(head(alcohol)) to show me the first six rows of the dataset. Copy and paste your code and output into the answer box.<\/p>\n
Print(head(alcohol))<\/p>\n
Question 2<\/p>\n
unemrate is the unemployment rate in a respondent\u2019s State. At what level of measurement is this variable? (Nominal Scale, Ordinal Scale, Interval Scale, or Ratio Scale)?<\/p>\n
Ratio scale since it has a defined zero point.<\/p>\n
Question 3-6<\/p>\n
The famsize variable measures how many people are in the respondent\u2019s family. Fill in the missing values in the following relative frequency table: <\/p>\n
Family Size Number of Respondents Relative Frequency Cumulative Relative Frequency <\/p>\n
1 2595 0.264 0.264 <\/p>\n
2 2311 0.235 0.499 <\/p>\n
3 1766 0.180 0.679 <\/p>\n
4 2000 0.204 0.883 <\/p>\n
5 740 0.075 0.958 <\/p>\n
6 257 0.026 0.984 <\/p>\n
7 99 0.010 0.995 <\/p>\n
8 32 0.003 0.998 <\/p>\n
9 9 0.001 0.999 <\/p>\n
10 6 0.001 0.999 <\/p>\n
11 5 0.001 1.000 <\/p>\n
12 1 0.000 1.000 <\/p>\n
13 1 0.000 1.000 <\/p>\n
Total number of respondents = 9822<\/p>\n
Family size 2: 2311\/9822 = 0.235<\/p>\n
Family size 4: 2000\/9822 = 0.204<\/p>\n
Cumulative Relative Frequency Family size 3: 0.499+0.180 = 0.679 <\/p>\n
Cumulative Relative Frequency Family size 3: 0.679+0.204 = 0.883<\/p>\n
Question 7-8f<\/p>\n
Create a barplot of the famsize variable Copy and paste your code in the answer box. In the next question, upload your barplot.<\/p>\n
# Simple Bar Plotcounts <- table(mtfamily$size)barplot (counts, main=”Number of Respondents”,\u00a0\u00a0 xlab=”Family Size”<\/p>\n
Question 9<\/p>\n
Does family size appear to be right-skewed, left-skewed, or not skewed at all?<\/p>\n
Family size appears to be left-skewed<\/p>\n
Question 10-11<\/p>\n
educ is the respondent\u2019s level of education, in years. Calculate the sample mean and sample standard deviation of educ.<\/p>\n
Sample mean = average (E2:E9823) = 13.30961<\/p>\n
Sample standard deviation =std. s (E2:E9823) = 2.898751<\/p>\n
Question 12<\/p>\n
Suppose we assume that unemrate is normally distributed with mean 5.57% and standard deviation 1.51%. What is the probability of picking a person at random who faces an unemployment rate less than 4%?<\/p>\n
\u00b5 = 0.0557 and \u03c3 = 0.0151<\/p>\n
P(x<0.04) = P(x-\u00b5<0.04-0.0557) = P (x-\u00b5\u03c3<0.04-0.05570.0151)0.04-0.05570.0151= -1.04P(z<-1.04) = 0.1492<\/p>\n
Question 13<\/p>\n
What is the probability of picking a person who faces an unemployment rate greater than 7%, assuming the unemployment rate is normally distributed as in question 12?<\/p>\n
\u00b5 = 0.0557 and \u03c3 = 0.0151<\/p>\n
P(x>0.07) = P(x-\u00b5>0.07-0.0557) = P (x-\u00b5\u03c3>0.07-0.05570.0151)0.07-0.05570.0151= 0.95P(z>0.95) = 0.1711<\/p>\n
Question 14<\/p>\n
What is the 98th percentile of unemrate, assuming it is normally distributed as in question 8?<\/p>\n
0.98 to z score <\/p>\n
Percentile to z score<\/p>\n
P(z<?) = 0.98<\/p>\n
Z = 2.05<\/p>\n
2.05=x-0.05570.0151X = 2.05(0.0151) + 0.0557<\/p>\n
X = 0.0867 = 8.67%<\/p>\n
Question 15-16<\/p>\n
Now let\u2019s informally test our assumption that the unemployment rate is normally distributed.<\/p>\n
plot a histogram of unemrate, using a binwidth of 0.1. Copy and paste the code you used in the answer box. In the next question, upload your histogram as a .png or .jpeg<\/p>\n
Question 17<\/p>\n
Based on your histogram, do you think the unemployment rate is normally distributed? Justify your reasoning in 2-3 sentences.<\/p>\n
The graph is not normally distributed.<\/p>\n
The graph is nowhere close to the bell shape. It peaks at somewhere in the middle of the histogram and also further to the right.<\/p>\n
Question 18<\/p>\n
Suppose you are rolling a six-sided die. Let event A= {2,4,6} and B = {1,3,5}. True or False, these events are mutually exclusive.<\/p>\n
False. The two events can happen at the same time.<\/p>\n
Question 19<\/p>\n
Suppose you are rolling a six-sided die. Let event A= {2,3,4} and event B= {1,2,3}. What is the intersection of these two events? That is, what is A\u2229B?<\/p>\n
A\u2229B = {2,3}<\/p>\n
Question 20<\/p>\n
Suppose you are rolling a six-sided die. Let event A= {2,3,4} and event B= {1,2,3}. What is the union of these two events? That is, what is A\u222aB?<\/p>\n
A\u222aB = {1,2,3,4}<\/p>\n
Question 21<\/p>\n
Suppose you flip a biased coin two times. The probability of getting heads in this coin is p = 0.2. What is the probability of getting two heads in the two flips?<\/p>\n
Hint: There are two ways to solve this. You can use R\u2019s built-in Binomial distribution (type help(dbinom)) for more information) or you can write out each event and calculate the probability using the probability rules.<\/p>\n
Since the probability of getting heads in one toss\/flip = 0.2<\/p>\n
The probability of getting two heads on two-coin tosses\/flips = 0.2 x 0.2 (because these are independent events) = 0.04<\/p>\n
Thus, the answer is:\u00a00.04<\/p>\n
Question 22<\/p>\n
Suppose X is Bernoulli distributed with parameter p=0.8. What is the mean of X?<\/p>\n
Bernoulli distribution is a type of discrete probability distribution which have two possible outcomes where probability of x = 0(failure) is 1-p and probability of x = 1(success) is p.<\/p>\n
Mean = Summation xp(x) = 0 * (1-p) + 1 * (p) = p = 0.8<\/p>\n
Therefore, mean(x) = 0.8<\/p>\n
Question 23<\/p>\n
Suppose you know that the number of customers that arrive at a grocery store in an hour is a Poisson random variable with \u03bb=200 That is, you know that on average 200 customers enter the store every hour. On average, how many customers can you expect to arrive in the next ten minutes?<\/p>\n
Here \u03bb = 200\/hr<\/p>\n
Or, \u03bb = 2003 min=103Expected number of customers in the next minute = \u03bbt<\/p>\n
= 10310=1003=33 customers <\/p>\n","protected":false},"excerpt":{"rendered":"
Midterm Morgan Holland 5\/22\/2020 Instructions You will need alcohol. RData dataset to complete the Midterm. The dataset includes data on<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-52659","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\n