← Exercises for Ch 3
Alternative solution: In a normal distribution, the mean is half way between the given extremes: μ = (4.50+8.50)/2 = 6.50. Then the distance from the mean to 8.50 must be three SD: 8.50−6.50 = 2.00 = 3σ; σ = 0.67 ounces.
|Ages||Midpoint (L1)||Frequency (L2)|
|20 – 29||25||34|
|30 – 39||35||58|
|40 – 49||45||76|
|50 – 59||55||187|
|60 – 69||65||254|
|70 – 79||75||241|
|80 – 89||85||147|
Caution! The midpoints are not midway between lower and upper bounds, such as (20+29)/2 = 24.5. They are midway between successive lower bounds, such as (20+30)/2 = 25.
1-VarStats L1,L2 (Check n first!)
x̅ = 63.85656971 → x̅ = 63.86
s = 15.43533244 → s = 15.44
n = 997
Common mistake: People tend to run 1-VarStats L1, leaving off the L2, which just gives statistics of the seven numbers 25, 35, …, 85. Always check n first. If you check n and see that n = 7, you realize that can’t possibly be right since the frequencies obviously add up to more than 7. You fix your mistake and all is well.
(b) You need the original data to make a boxplot, and here you have only the grouped data. A boxplot of a grouped distribution doesn’t show the shape of the data set accurately, because only class midpoints are taken into account. The class midpoints are good enough for approximating the mean and SD of the data, but not the five-number summary that is pictured in the boxplot.
n = 14 (This is the number of credits attempted. If you get 5, you forgot to include L2 in the command.)
x̅ = 2.93
Commute Distance, km 0- 9 4 10-19 12 20-29 7 30-39 1 40-49 1 Total 25
|(b) The class width is 10 (not 9). The class
midpoints are 5, 15, 25, 35, 45 (not 4.5, 14.5, etc.).
(c) Class midpoints in one list such as L2
and frequencies in another list such as L3. This is a sample, so
symbols are x̅, s, n, not μ, σ, N.
(d) Data in a list such as L1. 1-VarStats L1 gives x̅ = 17.6 km, Median = 17, s = 9.0 km, n = 25
(f) Mean, because the data are
Or, median, because there is an
Comment: The stemplot made the data look skewed, but that was just an artifact of the choice of classes. The boxplot shows that the data are nearly symmetric, except for that outlier. This is why the mean and median are close together. This is a good illustration that sometimes there is no uniquely correct answer. It’s why your justification or explanation is an important part of your answer.
(g) The five-number summary, from MATH200A part 2 [
1, 12, 17 22.5, 45. There is
one outlier, 45.
(The five-number summary includes the actual min and max, whether they are outliers or not.)
Draw an auxiliary line at z = −2. You know that the area between z = −2 and z = +2 is 95%, so the area between z = 0 and z = 2 is half that, 47.5% or 0.475.
zJ = (2070−1500)/300 = 570/300 = 1.90
zM = (129−100)/15 = 29/15 = about 1.93
Because she has the higher z-score, according to the tests Maria is more intelligent.
Remark: The difference is very slight. Quite possibly, on another day Jacinto might do slightly better and Maria slightly worse, reversing their ranking.
|Test Scores||Frequencies, f
|Class Midpoints, x
Put class midpoints in a list, such as L1, and frequencies go in another list, such as L2. (Either label the columns with the lists you use, as I did here, or state them explicitly: “class marks in L1, frequencies in L2”.)
(Always write down the command that you used.)
(a) n = 154
(b) x̅ = 499.81 (before rounding, 499.8051948)
(c) s = 12.74 (before rounding, 12.74284519)
Be careful with symbols. Use the correct one for symbol or population, whichever you have.
Common mistake: The SD is 12.74 (Sx), not 12.70 (σ), because this is a sample and not the population.