Intro
A student selects their roommates as the sample to represent the population of their dorm hallway. This is known as ______ sampling.
Convenience
Consider the random sampling of Iowa student to determine how often they eat at Micky's which is recorded as "one or more a week," "once or more a month," "once or more a year," and "never." The kind of variable being used here is a ______ variable.
Categorical
True or False:
1. P(A union B) = P(A intersect B)
2. P(A | B) = P(B | A)
1. False, P(A union B) = P(A) + P(B) - P(A intersect B)
2. False, P(A|B) is the probability of A occurring given B is true while P(B|A) is the probability of B occurring given A is true.
A large high school has 1,200 students evenly distributed across four grade levels. A researcher decides to randomly select 50 students from each grade level and survey them.
1. What type of sampling method is being used?
2. If the researcher instead randomly selected 8 entire homeroom classes and surveys every student in those homerooms, what type of sampling method is being used?
1. Stratified (strata = grades, randomly select from each grade)
2. Cluster (clusters = homerooms, randomly select entire homerooms)
A dataset has a mean of 72, median of 75, and standard deviation of 4. What is the shape of the distribution, and which measure of central tendency would be most appropriate to report?
Left-tailed (mean < median)
The median is the most appropriate (resistant to outliers)
Consider these events to be independent:
1. You roll a six-sided die
2. You flip a coin
What is the probability of rolling a 6 followed by a heads?
1/6 * 1/2 = 1/12
Matt is interested in the heights of students at the University of Iowa. In his sample, he found a sample variance of 9 inches. What is the sample standard deviation?
3
sqrt(9) = 3
s2 = sample variance
s = sample standard deviation
On a quiz, the mean score was 70 with a standard deviation of 5. Maria scored an 82. Calculate Maria's z-score and interpret what it means in context.
(82 - 70) / 5 = 2.4
Maria scored 2.4 standard deviations above the mean.
The probability that George gets an A in a class is 0.5, whereas the probability Brayton gets an A in that class is 0.6. The probability they both get an A is 0.3. What is the probability that neither of them get an A?
0.2
= 1 - 0.2 - 0.3 - 0.3
The average time spent studying per week by all University of Iowa students is 6 hours. The population variance is 0.3 hours, and the population standard deviation is 0.9 hours. You conduct a simple random sample of University of Iowa students and discover the sample's average weekly studying time to be 4.8 hours with a standard deviation of 0.5 hours.
Calculate (x_bar - mu) / s2
= (4.8 - 6) / 0.52
The five-number summary for a dataset is:
- Min = 12
- Q1 = 20
- Median = 25
- Q3 = 30
- Max = 60
Determine whether the maximum value is an outlier or extreme outlier.
IQR = 30 - 20 = 10
30 + 3*10 = 60, so the maximum value is an extreme outlier.
The probability that Matt hits a dartboard is 0.21. The probability Mikey hits the dartboard is 0.68. The probability that either Matt or Mikey hits the dartboard is 0.80. What is the probability that Matt hits the dartboard given that Mikey hits the dartboard?
0.1324
0.80 = 0.21 + 0.68 - P(M intersect K)
P(M intersect K) = 0.09
P(M | K) = 0.09 / 0.68 = 0.1324
{70, 115, 100, 85, 130}
Calculate the sample variance.
562.5
x_bar = (70+115+100+85+130) / 5 = 100
s2 = (70 - 100)2 + (115 - 100)2 + (100 - 100)2 + (85 - 100)2 + (130 - 100)2 / (5 - 1)
= 562.5
A student tracked the number of text messages sent each day for five days. The data are represented as deviations from the mean:
∑(x−x_bar)=0
Four of the daily values are known:
12, 18, 20, 15
The fifth day’s value is unknown. If x_bar = 18, use summation to find the missing fifth value.
(12 - 18) + (18 - 18) + (20 - 18) + (15 - 18) + (x - 18) = 0
-6 + 0 + 2 - 3 + x - 18 = 0
x - 25 = 0
x = 25
A rare disease affects 5% of a population. A new diagnostic test has been developed to detect this disease. However, the test is not perfect. If a person has the disease, the probability the test correctly detects it is 95%. If a person does NOT have the disease, the probability that the test correctly returns negative is 90%.
Suppose a person is randomly selected from the population, and they test positive for the disease. What is the probability this person actually has the disease?
0.33
P(D) = 0.05 P(~D) = 0.95
P(T+|D) = 0.95 P(T+|-D) = 1 - .90 = 0.10
P(T+) = P(T+|D)*P(D) + P(T+|-D)*P(-D)
P(T+) = 0.95*0.05 + 0.10*0.95 = 0.1425
P(D|T+) = P(T+|D)*P(D) / P(T+)
P(D|T+) = (0.95)*(0.05) / 0.1425 = 0.3333