Scatterplots
Suppose the line of best fit for a scatter plot is y = 3.5x + 56
R^2 = 0.8836
Find R.
What is the residual for the data point (5, 74)
R = sqrt(0.8836) = 0.94
Residual : y = 3.5(5) +56 = 73.5
74 - 73.5 = 0.5
Find the 5 number summary for :
3, 5, 7, 8, 9, 10, 12, 14, 15
Min:
Q1:
Median:
Q3:
Max:
IQR:
Are there any outliers?
Min : 3
Q1: (5+7)/2 = 6
Median: 9
Q3: (12+14)/2 = 13
Max: 15
IQR: 13 - 6 = 7
6 - (1.5)(7) = -4.5
13 + (1.5)(7) = 23.5
No outliers
A hardware store currently has 100 light bulbs in stock, which it sells in packs of 5 bulbs. Let x = the number of packs purchased by a randomly selected customer. The probability distribution of x is:
x (packs) 1 2 3 4
P(x) 0.25 0.35 ? 0.15
What is P(x = 3)?
Calculate the expected value of x
If the store sells to 10 customers, how many packs should they expect to sell? Do they have enough lightbulbs in stock?
P(x = 3) : 1 - (0.25 + 0.35 +0.15) = 0.25
Expected value: 1(0.25) + 2(0.35) + 3(0.25) + 4(0.15) = 2.30
10(2.30) = 23 packs
23(5) = 115 > 100 Not enough in stock
A college surveyed 10,000 students about their preferences for morning classes (M) and evening classes (E). Results showed:
- 6000 students prefer morning classes
- 4500 students prefer evening classes
- 2000 students prefer both
Fill out the contingency table
1. How many students prefer evening classes but not morning classes?
2. What is the probability that a randomly selected student prefers morning classes but not evening classes?
3. What is the probability that a student prefers evening classes given that they prefer morning classes?
4. Are preferring morning classes and preferring evening classes independent events?
1. 2500
2. 4000/10000 = 0.40
3. P(E|M) = (P(M and E)) / P(M) = 2000/6000 = 0.33
4. P(M) * P(E) = P(M and E) : 6000/10000 * 4500/10000 = 2000/10000
0.60 * 0.45 = 0.20
0.27 NOT EQUAL 0.20
NOT INDEPENDENT
A smoothie shop uses a machine to automatically dispense strawberry juice into bottles. The amount dispensed, x, is normally distributed with:
Mean : 16 ounces
Standard Deviation : 0.25 ounces
1. What is the probability that a randomly selected bottle contains between 15.8 ounces and 16.3 ounces?
2. What amount of juice corresponds to the 90th percentile?
3. If two bottles are filled independently, what is the probability that both contain at most 16.1 ounces.
1. normaldist(16, 0.25) : P(15.8 < x < 16.3) = 0.673
2. inversecdf(normaldist(16, 0.25), 0.90) = 16.32
3. normaldist(15, 0.25) : P(-infinity < x < 16.1) = 0.6554
(0.65540)^2 = 0.4295
Suppose the line of best fit for a scatter plot is y = 1.1x + 52
R^2 = 0.9409
Find R
Find the residual for (30, 83)
R = sqrt(0.9409) = 0.97
Residual : y = 1.1(30) + 52 = 85
83 - 85 = -2
Find the 5 number summary for :
18, 25, 12, 32, 40, 22, 28, 35
Min:
Q1:
Median:
Q3:
Max:
IQR:
Any outliers?
Min : 12
Q1: (25+12)/2 = 18.5
Median: (32+ 40)/2 = 36
Q3: (22+28)/2 = 25
Max : 40
IQR : 25 - 18.5 = 6.5
18.5 - 1.5(6.5) = 8.75
25 + 1.5(6.5) = 34.75
No outliers
A pet store sells fish food in small containers. Let x = the number of containers a randomly chosen customer buys. The probability distribution of x is:
x (containers) 1 2 3 4
P(x) 0.20 0.30 0.25 ?
What is P(x = 4)?
Find the expected value of x
If 12 customers come in, how many containers of fish food should the store expect to sell?
P(x = 4) = 1- (0.20 + 0.30 + 0.25) = 0.25
Expected Value: 1(0.20) + 2(0.30) + 3(0.25) + 4(0.25) = 2.55
12(2.55) = 30.6
A factory produces digital thermometers that are supposed to read exactly 98.6 degrees when measuring human body temperature. However, due to small variations in manufacturing, the actual reading x from a randomly chosen thermometer follows a normal distribution with Mean of 98.6 degrees and a standard deviation of 0.4 degrees.
1. What is the probability that a randomly selected thermometer reads between 98.2 and 99 degrees?
2. What reading corresponds to the 95th percentile?
3. If 3 thermometers are randomly selected, what is the probability all three read less than 99.1?
1. normaldist(98.6, 0.4) , P(98.2 < x < 99) = 0.683
2. inversecdf(normaldist(98.6, 0.4), 0.95) = 99.26
3. normaldist(98.6, 0.4) , P(-infinity < x < 99.1) = 0.894
(0.894)^3 = 0.7145