An Overview of Body Mass Index

Midterm
Don’t leave anything blank. I can always find a way to give you some credit if you’ve made an
attempt. Be sure to post any questions you have after the answers open. Unlike the word
problems in your book, my word problems are not real.
The work you turn in by 11:59 P.M. on Feb 16th must be your own. Please don’t get advice or
suggestions from your spouse, child, friend, or co-worker who just finished a similar class. If you
submit early in the week, I’m happy to provide that advice and those suggestions and then you
can resubmit, if you choose, so there is no reason to cheat. Be sure to show some work or tell me
what technology you used if you don’t type out equations and show step by step explanations.
1. (one point) A researcher collected BMI (Body Mass Index) of 50 students in an
Engineering school. He found the mean BMI was much smaller than the median BMI.
The reason for this might be___
a. Several students had much smaller BMI scores than the other students
b. Several students had much larger BMI scores than the rest
c. The researcher made some measurement errors
d. All the students had smaller BMI scores, compared to national average
Ans: Option b is correct as in the sample median is more than mean or the distribution is negative
skewed or majority of the sample have more BMI scores than the rest. Also, option c is possible as there
is always a chance that researcher made some measurement errors.
2. (one point) In the graph, there are two distributions: A and B.
We can know from the graph:
a. Distribution A has a bigger standard deviation than distribution B

b. Distribution B has a bigger standard deviation than distribution A
c. Distribution A and distribution B have the same standard deviation
d. We cannot tell which has a larger standard deviation from the graph
Ans: From the graph it is evident that the spread of distribution A is lower that spread of B as the range
of A is 50-10 = 40 and range of B = 90-10 = 80. As standard deviation is a measure of spread about the
mean hence option (b) is correct.
3. (one point) What is the level of measure for each of the following:
a. Final grades in this class (letter on transcript) __________Ordinal__________
b. Colors of Skittles ________Nominal____________
c. Amount in my bank account _______Numeric______________
d. Freezing temperatures on Celsius scale _________Numeric____________
4. one point) Determine type of sampling used:
a. I collected shoe sizes from my family _________Convenience
sampling(Non-probabilistic sampling)____________
b. 1003 adults were called after their phone
numbers were randomly generated by computer ________Simple random
sampling____________
c. Police stopped every sixth driver for sobriety test __________systematic random
sampling__________
d. U.S.D.O.C. collected data by randomly selecting
federal prisons and surveying every prisoner __________stratified
sampling___________
5. (two points) A simple random sample of pages of my dictionary was collected. These are
the number of words defined on each page. Find the measures of central tendency: mean,
median, and mode.
51 63 63 43 43 63 69 39 53 63
Mean = (51 + 63 + 63 + 43 + 43 + 63 + 69 + 39 + 53 + 63)/10 = 55
Median = Middlemost value of data in ordered manner = middle value of 53 and 63 as sample
size is even = 58.
Mode = value with highest number of occurrence = 63

There are 1549 pages. Estimate the total number of words in the dictionary. Is the
estimate likely to be an accurate estimate of the total number of words in the English
language?
The average number of words in the sample of size 10 is estimated to be 55. Now, assuming the
sample mean is equal to the population mean the total number of words in the dictionary is
Total words = total number of pages * average number of words in a page = 1549*55 = 85195
words. Now, by the central limit theorem if random samples of sufficient large size (n>=30) is
collected from a population with mean μ and standard deviation of σ, then the mean of sampling
distribution is equal to the population mean and sample standard deviation equal to σ/sqrt(n).
Thus as the sample size increases the sample mean goes close to the population mean. Now, in
this case only 10 pages are selected which means sample size 10 which is relatively low. Hence,
the estimate of population mean has more error and hence estimate of the total number of words
in the English language is not much accurate.
6. (two points) The bi-monthly salaries for the 12 workers at my local Chick Fil-A are as
follows:
Worker 1 and
7
Worker 2, 8 & 9 Worker 3 Worker 4, 10, 11, 12 Worker
5
Worker 6
774 665 1110 586 591 612
Calculate the measures of variability: range, variance, and standard deviation. Create a frequency
distribution table that includes relative and cumulative relative frequency. Then create a graph of
your choosing.
The salaries of 12 workers is loaded in excel and the measures of variability is obtained.
range 524
variance 22818.42
standard
deviation
151.0577
Bi-monthly
salary
Frequency Relative frequency
(%)
Cumulative relative frequency (%)
586 4 33.33333333 33.33333333
591 1 8.333333333 41.66666667
612 1 8.333333333 50
665 3 25 75
774 2 16.66666667 91.66666667

1110 1 8.333333333 100
Total 12 100
Bar chart of relative frequency distribution:
7. (two points) I had 14 students last term who earned the following grades:
98 85 89 73 72 85 85 92 85 97 92 93 95 86
What is the 5 numbers summary for these students? Create a box plot for the data.
Five numbers summary:
minimum 72
1st
quartile
85
median 89
3rd
quartile
94
maximum 98

100
95
90
85
80
75
70
data
Boxplot of data
8. (two points) I developed an achievement test and normalized it on people throughout the
country over a year. My test has scores that are normally distributed with a mean of 100 and a
standard deviation of 20.
a. Find the probability that a randomly selected adult has IQ < 115
b. Find the probability that a randomly selected adult has IQ > 70
c. Find P30, which is the IQ score at the 305h percentile.
d. What percentage of adults have scores between 80 and 120?
Ans: Given, μ = 100 and σ = 20
a. Now, P(IQ<115) = P(Z<(115-100)/20) = P(Z< 0.75) = 0.7734 or 77.34%.
b. P(IQ>70) = 1 – P(IQ<=70) = 1 – P(Z<=-1.5) = 1 – 0.066807 = 0.9332 or 93.32%.
9. (two points) Ten peas are generated from parents having the green/yellow pair of genes so
there is a 75% probability that an individual pea will have a green pod. Find the probability that
among the 10 offspring peas, no more than 5 have green pods.
Ans: Let the event of being a green pod is G.
Given, P(G) = p = 0.75, P(G’) =q = 1- 0.75 = 0.25.
Total number of peas generated (n) = 10

Hence, assuming G follows binomial distribution with p= 0.75 and q = 0.25 the probability there
will be no more than 5 green pods out of 10 pods is
P(G<=5) = P(G= 0) + P(G=1) = P(G=2) + P(G=3) + P(G=4) + P(G=5)
= 10C0 * (0.75)^0 * (0.25)^10 + 10C1 * (0.75)^1 * (0.25)^9 + 10C2 * (0.75)^2 * (0.25)^8 +
10C3 * (0.75)^3 * (0.25)^7 + 10C4 * (0.75)^4 * (0.25)^6 + 10C5 * (0.75)^5 * (0.25)^5 = 0.0781
or 7.81%.
10. (two points) I always wanted to give away my vast wealth, so I am going to have a lottery
next year. You have to pick the five winning numbers from 1-25. Each randomly selected
number is different and order doesn’t matter. What is the probability of your winning?
Here total numbers = 25.
Winning numbers = 5.
Now, 5 numbers can be chosen from 25 numbers in 25C5 different ways.
Number of ways 5 winning numbers are chosen = 1 way as order does not matter.
Hence, probability of winning = 1/(25C5) = 1/53130.
11. (two points) In a study of families with children with disabilities, groups of 6 US households
were randomly selected. In the table below, the random variable x represents the number of
households among 6 that had a child with a disability living there.
x P(x)
0 0.02
1 0.15
2 0.29
3 0.31
4 0.21
5 0.02
a) Is this a probability distribution?
As P(xi>=0) for all xi
And ∑ P( xi) = 1
Hence, the x is a probability distribution.

b) What is the mean?
The mean is E(x) = ¿ = 2.6/1 = 2.6.
c) What is the standard deviation?
Standard deviation = √Var ( x)
Now, Var (x ) = E( x2 )– E2 ( x) = 26.78 – 2.6^2 = 20.02.
Hence, Standard deviation = 4.474
12. (two points) I am going to roll two fair dice.
a) List the sample space
Sample space size = 6^2 = 36.
Sample space = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)
(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)
(6,4)(6,5)(6,6)}
b) What is the probability that I will roll a 3 on one of them when I roll both at the same
time?
Require probability = P(rolling 3 on one die) + P(rolling 3 on other die)
= (1/6)*(5/6) + (1/6)*(5/6) = 10/36 = 5/18.
c) What is the probability that I will roll at least a sum of 7 when I roll them both?
P(rolling at least sum of 7) = 1 – P(rolling at most sum of 7) = 1 – (P(rolling exactly sum
of 2) + P(rolling exactly sum of 3) + P(rolling exactly sum of 4) + P(rolling exactly sum
of 5) + P(exactly sum of 6) + P(exactly sum of 7))
P(rolling exactly sum of 2) = {(1,1)} = 1 case
P(rolling exactly sum of 3) = {(1,2),(2,1)} = 2 cases
P(rolling exactly sum of 4) = {(2,2),(1,3),(3,1)} = 3 cases
P(rolling exactly sum of 5) = {(1,4),(4,1),(2,3),(3,2)} = 4 cases
P(rolling exactly sum of 6) = {(1,5),(5,1),(2,4),(4,2),(3,3)} = 5 cases
Set that gives the sum of 7 when two dice are rolled simultaneously = {(1,6),(6,1),(2,5),
(5,2),(3,4),(4,3)} = 6 cases
Hence, P(rolling at least sum of 7) = 1 – (1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36) = 1 –
21/36 = 15/36.
d) Define mutually exclusive
N events X1,X2,X3,…XN are called mutually exclusive if occurrence of one indicates
non-occurrence of other and P(X1 U X2 U X3 … U XN) = 1
e) Define independent
N events X1,X2,X3,…XN are called independent or mutually independent if occurrence
of one does not depend on occurrence of others giving the relation P(X1 ∩ X2 ∩ X3 ∩…
XN) = P(X1)*P(X2)*P(X3)*…P(XN)