BST Numericals
BST Numericals
BST Numericals
Class Discussions
(Numerical Questions)
Examples of Arithmetic Mean
The following table shows monthly sales of 10 The number of LED lamps used in households is
stores of a retail chain. Calculate the average given below. Calculate the average number of LED
monthly sales. lamps used.
(Ungrouped) (Grouped, discrete)
Sales Average Sales No. of LED No. of
Store Average LED
($ 1000s) calculation lamps households f*X
calculation
A 22 (X) (f)
B 25 Average monthly sales Average no.
1 2 2*1=2
C 27 = of LED
(22+25+27+29+30+31 lamps used
D 29 2 4 4*2=8
+32+33+35+36)/10 = ΣfX/Σf
E 30 = ΣX/n = 60/20
F 31 = 300/10 3 6 6*3=18 =3
= 30 The average
G 32
The average monthly 4 8 8*4=32 usage of
H 33 sales is $30,000 per LED is 3
I 35 store. lamps per
Total Σf = 20 ΣfX = 60
J 36 household.
The following table provides data on the distribution of salaries ($1000s)
of 50 employees of an organization. Calculate the average salary.
(Grouped, continuous)
No. of
Salary Average Salary
Employees Mid-value (X) f*X
($1000s) calculation
(f)
10-20 2 15 2*15=30
20-30 4 25 4*25=100 Average salary
30-40 6 35 6*35=210 of employees
= ΣfX/Σf
40-50 8 45 8*45=360 = 2750/50
50-60 10 55 10*55=550 = 55
1. The mean weights of five computer stations is 167.2 lbs. The weights of four of
them are 158.4 lbs, 162.8 lbs, 165.0 lbs and 178.2 lbs respectively. What is the
weight of the fifth computer?
2. The following table gives the weights of wooden items being sold by a timber
merchant. Calculate mean weight of the items sold.
Weight (lbs) 1-3 4-6 7-9 10-12 13-15
No. of items 8 25 45 18 4
3. An ice-cream parlor sells six varieties of ice-creams which have generated the
following revenue. Find the mean price of an ice-cream sold.
Ice-cream Butter scotch Chocolate Lychee Choco chips Tooty fruity Vanilla
Price (Rs.) 40 90 65 55 75 45
Sales (Rs.) 5,00,000 4,50,000 3,38,000 3,01,180 4,93,800 3,14,415
Weighted Arithmetic Mean
The weighted mean is a type of mean that is calculated by multiplying the weight (or
probability) associated with a particular event or outcome with its associated quantitative
outcome and then summing all the products together.
Weighted Mean = ΣWiXi/ΣWi; i = 1,2,3,……,n.
Weight
Examination Score (X) W*X Weighted Mean
(W)
2. A batsman scored 1, 113, 148, 22, 24, 27, 15, 16, 16 & 28 runs in the last 10 innings. Using an
appropriate measure, find his average score.
Key: Since there are 2 extreme scores 113 & 148, hence mean would be affected by these values.
Here, median would be an appropriate measure.
Arrangement: 1, 15, 16, 16, 22, 24, 27, 28, 113, 148.
No. of observations, n = 10 (even)
Median = Mean of (n/2)th and (n/2+1)th observations
= Mean of 5th and 6th observations
= (22+24)/2 = 23
The average score of the batsman is 23 runs.
Example of Median (Grouped data, discrete series):
Calculate median marks obtained by the students:
Marks : 45 55 25 35 5 15
No. of students : 40 30 30 50 10 20
Key: Arrange the marks in ascending/descending order, then find CF.
Marks No. of students Cumulative Finding Median
(X) (f) frequency (CF) value
5 10 10 N/2 = 90
15 20 30 Since 90 is not
available in CF
25 30 60 column, we
35 50 110 take the next
value 110.
45 40 150 Hence the
55 30 180 (N) median marks
is 35.
Example: Grouped data, continuous series
Mode
• Mode is the value which occurs most frequently in a distribution.
• A distribution can have one or more than one modes.
• Mode is widely used while compiling the results of surveys. The options with
maximum frequencies are considered and decisions are taken accordingly.
• Mode can be calculated for grouped, ungrouped, discrete and continuous data.
• Mode for grouped and continuous series is calculated using the formula:
Discrete series Continuous series
Practice Problems
Range
• Range is defined as the difference between largest observation (L) and smallest
observation (S) in a distribution. Range = L-S
• If the average of two distributions are almost same, then the distribution with
smaller range is said to have less dispersion.
• Lesser value of range indicates more consistency in the distribution.
• Coefficient of range = (L-S)/(L+S)
• Range is widely used for statistical quality control. If the dimensions of products
are beyond a defined range, they are discarded.
• It facilitates the study of variations in the prices of shares, agricultural products
and other commodities.
• It also helps in weather forecasts by indicating minimum and maximum
temperature.
Quartile Deviation
• Quartile deviation or semi inter quartile range is half of the difference between upper
quartile (Q3) and lower quartile (Q1).
• Quartile deviation (QD) = (Q3 - Q1)/2, and Coefficient of QD = (Q3 - Q1)/(Q3 + Q1)
• The Quartile Deviation doesn’t take into consideration the extreme points of the
distribution. Thus, the dispersion or the spread of only the central 50% data is considered.
• It is the best measure of dispersion for open-ended systems (which have open-ended
extreme ranges).
• Q1 = [(n+1)/4]th observation, and Q3 = [3(n+1)/4]th observation
• Ex: 15, 16, 17, 18, 19, 20, 22, 26, 27, 30, 31, 34, 41, 45, 47, 52, 53, 56, 67, 68, 69, 72, 74,
76
Q1 = {(24+1)/4}th obs. = 6.25th obs. = 7th obs. = 22
Q3 = {3(24+1)/4}th obs. = 18.75th obs. = 19th obs. = 67
QD = (67 – 22)/2 = 22.5 and
Coeff. of QD = (67 – 22)/(67 + 22) =
Mean Deviation
• Mean deviation is arithmetic mean of the absolute deviations of all items from a
measure of central tendency.
• Mean deviation (MD) =
where ‘m(X)’ is any central tendency value and ‘n’ is no. of observations.
Example: Calculate Mean X |X-m(X)| Mean Deviation
7 |7-9|= 2
8 |8-9|= 1
13 |13-9|= 4
Mean
Σ|X-m(X)|= 14
m(X) = 9
Variance, Standard Deviation
2
x (x-𝑥)ҧ 𝑥 − 𝑥ҧ Variance Std. Deviation
8 8-7=1 1
4 4-7=-3 9
9 9-7=2 4 Sample variance (s2) Sample Std. Deviation (s)
11 11-7=4 16 = Σ 𝑥 − 𝑥ҧ 2 /(n-1) = Sqrt (Sample variance)
= 46/4 = Sqrt (11.5)
3 3-7=-4 16 = 11.5 = 3.391
Σx=35 2
Σ 𝑥 − 𝑥ҧ =
𝑥ҧ = 7
46
2 2
x f fx (x-𝑥)ҧ 𝑥 − 𝑥ҧ 𝑓 𝑥 − 𝑥ҧ Variance Std. Deviation
8 3 24 8-8=0 0 3*0=0
4 4 16 4-8=-4 16 4*16=64
10 6 60 10-8=2 4 6*4=24 Sample Variance (s2) Std Deviation (𝑠)
12 4 48 12-8=4 16 4*16=64 = Σf 𝑥 − 𝑥ҧ 2 /(Σf -1) = Sqrt (Sample variance)
= 200/19 = Sqrt (10.526)
4 3 12 4-8=-4 16 3*16=48 = 10.526 = 3.244
2
Σfx = 160; Σf=20; Σ𝑓 𝑥 − 𝑥ҧ
𝑥ҧ = 160/20 = 8 = 200
Example: Grouped, continuous series
A showroom of cars displays its sales figures for the last 30 days. Calculate the mean no. of
cars sold per day and std. deviation.
No. of
Days 2 2
cars x fx (x-𝑥)ҧ 𝑥 − 𝑥ҧ 𝑓 𝑥 − 𝑥ҧ Variance Std. Deviation
(f)
sold
0-2 14 1 14 1-3=-2 4 14*4=56
2-4 7 3 21 3-3=0 0 7*0=0
4-6 5 5 25 5-3=2 4 5*4=20 Sample Standard
Sample Variance (s2)
Deviation (s)
6-8 3 7 21 7-3=4 16 3*16=48 = Σf 𝑥 − 𝑥ҧ 2 /(Σf-1)
= Sqrt (Variance)
= 160/29
8-10 1 9 9 9-3=6 36 1*36=36 = Sqrt (5.517)
= 5.517
= 2.349
2
Σfx = 90; Σf = 30 Σ𝑓 𝑥 − 𝑥ҧ
𝑥ҧ = 90/30 = 3 = 160
Coefficient of Variation
CV is used to study consistency whenever there is a comparison between two or
more datasets.
Two companies Dawson Suppliers and Clark Distributors deliver construction materials. The following data shows
days of delivery for both the companies on 8 occasions. Which company is more consistent in deliveries?
Dawson Clark 𝟐 𝟐
ഥ
𝒙−𝒙 ഥ
𝒙−𝒙 ഥ
𝒚−𝒚 ഥ
𝒚−𝒚 Coefficient of Variation
(x) (y)
11 8 1 1 -2 4
10 10 0 0 0 0
9 17 -1 1 7 49
10 7 0 0 -3 9 SD of x = Sqrt (26/7) = 1.92
8 10 -2 4 0 0 CV of x = (1.92/10)*100 = 19.2%
8 11 -2 4 1 1
SD of y = Sqrt (72/7)= 3.21
10 10 0 0 0 0 CV of y = (3.21/10)*100 = 32.1%
14 7 4 16 -3 9
Dawson Suppliers is more consistent in
2 2 delivering the materials.
Mean Mean Σ 𝑥 − 𝑥ҧ Σ 𝑦 − 𝑦ത
𝑥ҧ = 10 𝑦ത = 10 = 26 = 72
Practice Problems
1. Find quartile deviation from the following data:
109, 189, 167, 209, 309, 265, 189, 187, 165, 239, 308, 378, 367, 109, 198, 209, 218, 387
2. The share prices of two companies X and Y are given below for twelve days. Which
company’s share prices are more consistent?
Days 1 2 3 4 5 6 7 8 9 10 11 12
X 201 200 199 203 206 208 206 201 197 199 198 196
Y 291 293 293 287 292 298 298 299 302 302 302 304
When computing the z-score for each sample on the data set a threshold must be specified. A general
‘thumb-rule’ for detecting outliers is lZl > 3.0, however it varies. Sometimes lZl values more than 2.5 are
also considered as outliers.
Meal price (x) ($) Z-score = (x-mean)/s.d.
18 -0.3545
19 -0.3151
20 -0.2757
17 -0.3939
21 -0.2363
22 -0.1969
99 2.8358 (Outlier)
21 -0.2363
15 -0.4726
18 -0.3545
Mean=27
S.D.=25.3859
‘Box-plot’ or ‘IQR’ method:
Interquartile range (IQR) is a measure of variability and also referred to as ‘midspread’. It is calculated as:
IQR = Q3 – Q1
With the help of IQR, we determine lower limit and upper limit for the dataset. Any value less than lower limit or
more than upper limit is considered as an outlier.
Lower limit = Q1 – 1.5*IQR
Upper limit = Q3 + 1.5*IQR
Meal price ($) Calculations
15 IQR = Q3 – Q1
17 Q1 = 18 and Q3 = 21
IQR = 3
18 Lower limit = 18 – 1.5*3 = 13.5
18 Upper limit = 21 + 1.5*3 = 25.5
19 Any value less than $13.5 and more than
20 $25.5 will be considered as outlier.
Hence the meal price of $99 is an outlier.
21
21
22
99
The data for ‘Stereo and Sound Equipment Store’ is given pertaining to the no. of television commercials shown and sales during
10 weeks. The manager of store wants to determine the association between television commercials and sales revenue. Calculate
sample covariance and interpret the result.
No. of
Week Commerci Sales ($100s) (x-ഥ
𝒙) (𝒚 − 𝒚
ഥ) (x-ഥ
𝒙)(𝒚 − 𝒚
ഥ)
als (x) (y)
Cov(X,Y) = Σ(x-𝑥)(𝑦
ҧ − 𝑦)/(n-1)
ത
1 2 50 -1 -1 1
= 99/(10-1)
2 5 57 2 6 12 = 11
9 4 59 1 8 8
10 2 46 -1 -5 5
where dx = X-A & dy = Y-B; A & B are assumed means from X & Y series respectively.
Ex: Find the Karl Pearson’s Coefficient of Correlation between X and Y from the following data:
X Y dx = X-16 dy = Y-27 (dx)2 (dy)2 dx.dy
10 19 -6 -8 36 64 48
12 22 -4 -5 16 25 20
13 26 -3 -1 9 1 3
16 27 0 0 0 0 0
17 29 1 2 1 4 2
20 33 4 6 16 36 24
25 37 9 10 81 100 90
∑X = 113 ∑Y = 193 ∑dx = 1 ∑dy = 4 ∑(dx)2 = 159 ∑(dy)2 = 230 ∑dx.dy = 187
7x187−1x4
r=
7x159 − 1x1 { 7x230 −(4x4)}
1309−4
=
(1113−1)(1610−16)
1305
=
(1112)(1594)
1305
=
1330.9
r = 0.986
There is very strong direct (or strong positive)
correlation between age and weight.
Practice Exercises
1. The sample gives data collected by department of transportation on driving speed (miles
per hour) and fuel economy (mileage per gallon) of mid-sized automobiles. Is there any
correlation between the two variables? Interpret your result.
Speed 30 50 40 55 30 25 60 25 50 55
Economy 28 25 25 23 30 32 21 35 26 25
2. A leading TV manufacturing company was trying to establish if there was any correlation
between family size and TV screen size from the following data:
Family size 2 3 4 3 2 4 2 4
TV size 32 28 40 48 32 28 28 28
YC 2 2 3 4 3 1 4 3 4 4
AI 20 23 25 26 28 29 27 30 33 35
Q2. A company sells fitness equipment in various price range. Ratings of users have been provided in the following data
along with price of the equipment ($100s):
Model A B C D E F G H
Price 37 25 28 19 10 8 17 6
Rating 87 84 82 74 73 69 68 55
Example-1: A card is drawn from a pack of 52 cards, find the probability that it is either a
queen or a heart.
Solution: Let A is the event that the card is a queen and B is the event that it is a heart.
Total no. of queens = 4 and total no. of hearts = 13, so P(A) = 4/52 and P(B) = 13/52
There is a card ‘queen of heart’ which includes both events A and B, P(A and B) = 1/52
Hence, P(A∪B)= P(A) + P(B) – P(A∩B)
= 4/52 + 13/52 – 1/52
= 16/52 or 0.3077
The required probability is 0.3077.
Example-2: In a bank, 50% of the accounts are current accounts, 30% are savings accounts
and rest are loan accounts. What is the probability that a randomly selected customer is
having either a current account or a savings account?
Solution: Let A, B and C are the events that a customer is having current, savings and loan
accounts respectively.
P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2
Since the events are mutually exclusive, hence the required probability is
P(A∪B) = P(A) + P(B) = 0.5 + 0.3 = 0.8
There is 80% chance that the randomly selected customer is having either a current or a
savings account.
Solution:
First we denote by E the event that the household watches the news on the
evening and M the event that the household watches the morning news.
From the question we have P(E)=0.7 and P(E and M)=0.5.
We have to calculate P(M|E).
We can use the equation of the multiplicative law as follows:
P(E and M) = P(E)×P(M|E)
Hence, P(M|E) = P(E and M)/P(E)
P(M|E) = 0.5/0.7 = 0.714
Ex.1: A study of speed violations and drivers who use cell phones produced the following data:
Marginal probabilities:
Prob. that the driver uses a cell phone P(C) = 305/750 = 0.4067
Prob. that the does not use cell phone P(NC) = 445/750 = 0.5933
Prob. that the driver had speeding violation P(S) = 70/750 = 0.0933
Prob. That the driver did not have speeding violation P(NS) = 680/750 = 0.9067
Joint probabilities:
Prob. that the driver uses cell phones and had speed violation P(C∩S) = 25/750 = 0.0333
Prob. that the driver uses cell phones and had no speed violation P(C∩NS) = 280/750 = 0.3733
Prob. that the driver does not use cell phones and had speed violation P(NC∩S) = 45/750 = 0.06
Prob. that the driver does not use cell phones and had no speed violation P(NC∩NS) = 400/750 = 0.5333
Speed violations No speed violations Total Joint Probabilities
(S) (NS)
Uses cell phones while 25/750 = 0.0333 280/750 = 0.3733 305/750 = 0.4066
driving (C)
Does not use cell phones 45/750 = 0.06 400/750 = 0.5334 445/750 = 0.5934
while driving (NC)
Marginal Probabilities
Total 70/750 = 0.0933 680/750 = 0.9067 1
Conditional probabilities:
(i) Prob. that the driver uses a cell phone given that he had speed violation
P(C|S) = P(C∩S)/P(S) = 0.0333/0.0933 = 0.357
(ii) Prob. that the driver uses a cell phone given that he had did not have speed violation
P(C|NS) = P(C∩NS)/P(NS) = 0.3733/0.9067 = 0.4117
(iii) Prob. that the driver does not use a cell phone given that he had speed violation
P(NC|S) = P(NC∩S)/P(S) = 0.06/0.0933 = 0.6431
(iv) Prob. that the driver does not use a cell phone given that he did not have speed violation
P(NC|NS) = P(NC∩NS)/P(NS) = 0.5333/0.9067 = 0.5882
(v) Prob. that the driver had speed violation given that he uses cell phone
P(S|C) = P(C∩S)/P(C) = 0.0333/0.4067 = 0.0819
(vi) Prob. that the driver had speed violation given that he does not use cell phone
P(S|NC) = P(NC∩S)/P(NC) = 0.06/0.5933 = 0.1011
(vii) Prob. that the driver did not have speed violation given that he uses cell phone
P(NS|C) = P(C∩NS)/P(C) = 0.3733/0.4067 = 0.9179
(viii) Prob. that the driver did not have speed violation given that he does not use cell phone
P(NS|NC) = P(NC∩NS)/P(NC) = 0.5333/0.5933 = 0.8989
Ex.2: People use the cab services for personal and official purposes during weekdays and
weekends. The following data gives the usage pattern of 30,000 people:
Personal Official Total A. Joint and marginal probability table
purpose (C) purpose (D)
Personal Official Total
Weekdays 8000 14000 22000 purpose (C) purpose (D)
(A)
Weekdays (A) 0.2667 0.4667 0.7334
Weekends 6000 2000 8000
(B) Weekends (B) 0.2 0.0666 0.2666
Total 14000 16000 30000 Total 0.4667 0.5333 1
(i) What is the probability that delivery will be made within 3 to 6 days?
(ii) What is the probability that the delivery will be late?
(iii) What is the probability that the delivery will be early?
2. The following data shows the no. of hours a car being parked at a parking slot along with the
probabilities. The parking supervisor wants to know the expected no. of hours and standard deviation of
the no. of hours cars are parked in the slot.
No. of hours 1 2 3 4 5 6 7 8
Probability 0.24 0.18 0.13 0.10 0.07 0.04 0.04 0.20
Binomial Distribution
If there are only two possible outcomes of an experiment, say success and failure, then the
probability of ‘r’ successes in ‘n’ trials is given by:
P(X=r) = nCr pr (1-p)n-r ~ B(n, p)
where ‘p’ is the probability of success (which remains constant), r ≤ n and, r = 0, 1, 2, 3………n and
n = 1, 2, 3, 4, …………, ∞
✓ Sometimes it is also written as P(X=r) = nCr pr qn-r, where ‘q’ is the probability of failure and, p+q =
1.
✓ Mean and variance of a Binomial distribution are ‘np’ and ‘npq’ respectively.
Ex.1: Probability that a delivery boy delivers a correct order is 0.9. What is the probability that he
correctly delivers 8 out of 10 orders? What is the probability that he correctly delivers 8 or more
times?
Soln: Prob. of correct delivery (success) = 0.9; Prob. of wrong delivery (failure) = 1-0.9 = 0.1
n = 10, r = 8, p = 0.9, (1-p) = 1-0.9 = q
P(X=8) = 10C8 (0.9)8 (1-0.9)10-8 = 45.(0.4305).(0.01) = 0.1937
Hint: P(X≥8) = P(X=8) + P(X=9) + P(X=10); P(X=9) = 0.3874, P(X=10) = 0.3487; Answer: 0.9298
Ex.2: Nitrogen gas is filled in automobile tyres to improve the ride quality. A filling station experiences
that 30% of the customers get nitrogen gas filled in their vehicle’s tyres. If 10 customers arrive at a
gas station then what is the probability that,
(i) All of them would fill nitrogen gas?
(ii) None of them would fill nitrogen gas?
(iii) Half of them would fill nitrogen gas?
(iv) Less than 8 customers would fill nitrogen gas?
(v) Find the mean no. of customers who would fill nitrogen gas.
Soln: n = 10, p = 0.3, 1-p = 1-0.3 = 0.7
P(X=r) = nCr pr (1-p)n-r
(i) P(X=10) = 10C10 (0.3)10 (1-0.3)10-10 = 1.(0.000006).(1) = Almost zero
(ii) P(X=0) = 10C0 (0.3)0 (1-0.3)10-0 = 1.(1).(0.0282) = 0.0282
10!
(iii) P(X=5) = 10C5 (0.3)5 (1-0.3)10-5 = . (0.3)5(1-0.3)10-5 = 252.(0.0024).(0.1681) = 0.1029
5! 10−5 !
(iv) P(X<8) = 1-P(X≥8) = 1-[P(X=8) + P(X=9) + P(X=10)] = 1-(0.0015+0.0001+0.000006) = 0.9984
(v) Mean no. of customers who fill nitrogen gas = np = 10.(0.3) = 3
Ex.3: Fifty percent of the people believe that the country is in recession. For a sample of 20
people, make the following calculations:
(i) Probability that 12 people believe the country is in recession. (0.1201)
(ii) Probability that at least 18 people believe the country is in recession. (0.0002)
(iii) Probability that more than three people believe the country is in recession.
P(X>3)=1-P(X≤3) = 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)] = (0.9987)
(iv) Expected no. of people who would say that the country is in recession. (10)
(v) Compute variance and std. deviation of the no. of people who believe the country is in
recession. (5, 2.236)
Ex.4: Twenty three percent of the vehicles are not covered by any insurance. On a special
checking day, 30 vehicles are checked randomly. What is the probability that more than 27
vehicles are insured? (here, p=0.77, n=30)
What is the expected no. of vehicles not covered by any insurance? What is the variance and
std. deviation? (here, n=30, p=0.23, q=0.77)
Poisson Distribution
It is used to estimate the probability of no. of occurrences over a specified period of time. If the mean no. of
occurrences over a specified period of time is ‘λ’ then the probability of ‘r’ occurrences is given by,
−λ λ𝑟
P(X=r) = 𝑒 . ~ 𝑃(λ)
𝑟!
where r = 0, 1, 2, 3, …………∞ and e = 2.718.
✓ Both mean and variance of a Poisson distribution are ‘λ’.
✓ The value of ‘λ’ should be adjusted according to time-interval.
✓ Poisson distribution approximates the binomial distribution for large n and small p.
Ex.1: In a drive-thru window of a burger shop, average 10 customers arrive in a 30-minute interval. What is the
probability that exactly 5 customers arrive in 30 minutes?
What is the probability that 5 customers arrive in 15-minutes interval?
−λ λ𝑟
Soln: P(X=r) = 𝑒 . 𝑟!
For 30-minutes duration, λ = 10 and r = 5.
−10 105
P(X=5) = 𝑒 . = (0.000045).(100000/120) = (0.000045).(833.33) = 0.0378
5!
For 15-minutes duration, λ = 5 and r = 5.
−5 55
P(X=5) = 𝑒 . = (0.0067).(3125/120) = 0.1754
5!
Ex.2: An average of 15 aircraft accidents occur each year. Compute
(i) The mean no. of accidents per month.
(ii) Prob. of no accidents during a month.
(iii) Prob. of exactly one accident per month.
(iv) Prob. of more than one accidents per month.
− λ λ𝑟
Soln: P(X=r) = 𝑒 .
𝑟!
For 12-months duration λ = 15, for one-month duration λ = 1.25.
(i) Mean no. of accidents per month = 15/12 = 1.25
−1.25 (1.25)0
(ii) For r=0, P(X=0) = 𝑒 . = (0.2865).(1/1) = 0.2865
0!
−1.25 1.25 1
(iii) For r=1, P(X=1) = 𝑒 . = (0.2865).(1.25/1) = 0.3581
1!
(iv) For r>1, P(X>1) = 1- P(X≤1) = 1-[P(X=0)+P(X=1)] = 1-(0.2865+0.3581) = 0.3554
Ex.3: Phone calls arrive at the rate of 48 per hour in a call centre. Compute
(i) The mean no. of calls in 5 minutes duration. (4)
(ii) The prob. of receiving three calls in 5 minutes. (0.1952)
(iii) Prob. of receiving exactly 10 calls in 15 minutes. (0.1048)
(iv) Prob. of receiving at least one call in 10 minutes. (0.9997)
Ex.4: The safety department of Blue Pharma Company found that an average of three accidents occur per
month in the factory. Calculate the probability of (a) no accidents, (b) exactly one accident, (c) exactly two
accidents, (d) less than four accidents, (e) more than five accidents.
Soln: λ =3
− λ λ𝑟
P(X=r) = 𝑒 .
𝑟! 0
(3)
P(X=0) = 𝑒 −3 . = 0.0498, P(X=1) = 0.1494, P(X=2) = 0.2240
0!
P(X<4) = P(X=0)+P(X=1)+P(X=2)+P[X=3] = 0.0498+0.1494+0.2240+0.2240 = 0.6472
P(X>5) = 1-P[X≤5] = 1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008) = 0.0839
Ex.5: Patients arrive at a hospital at the rate of 6 per hour. Find the probability that in a 90-minute duration
(i) exactly 7 patients arrive in the hospital.
(ii) between 7 and 10 patients arrive in the hospital. P(7≤X≤10) = P(X=7)+P(X=8)+P(X=9)+P(X=10)
(iii) If a patient arrives at 11:30am then what is the probability that other patients arrive before 11:45am?
λ= 1.5, P(X≥1)= 1-P(X<1) = 1-P(X=0) = 1-e-1.5 = 1- 0.2232 = 0.7768
Normal Distribution: Practice Exercises
1. In a city, it is estimated that the maximum temperature is normally distributed with a
mean of 23°C and a standard deviation of 5°C. Calculate the number of days in this
month in which it is expected to reach a maximum of between 21°C and 27°C.
2. The mean weight of 500 college students is 70 kg and the standard deviation is 3 kg.
Assuming that the weight is normally distributed, determine how many students weigh:
a. between 60 kg and 75 kg
b. more than 90 kg
c. less than 64 kg
d. exactly 64 kg
e. 64 kg or more
Practice Exercises
3. For borrowers with good credit scores, the mean debt amount is $15,000. Assuming
the debt amounts to be normally distributed with standard deviation $3000, calculate
the probability that
a. debt for a borrower is more than $18,000
b. debt for a borrower is less than $10,000
c. Debt for a borrower is between $12,000 and $18,000
4. At a gas station, the daily demand for regular gasoline is normally distributed with a
mean of 1000 gallons and a standard deviation of 100 gallons. The station manager has
just opened the station for business and finds that there is exactly 1100 gallons of
regular gasoline in storage. The manager wants to know the chances that the available
gasoline is enough to satisfy the day’s demand.
Sampling Distribution: Practice Exercises
1. Mean expenditure of all the visitors in a restaurant is Rs.2000 with a std. deviation of
Rs.250. A random sample of 40 customers was taken, find the probability that
(a) mean expenditure of customers is more than Rs.1928, (b) mean expenditure of
customers is between Rs.1950 and Rs.2030.
(a) Z = Xσ −μ = 1928
250
−2000
= -1.82
ൗ n ൗ 40
Z σ 2 2
(1.645) (45) 2 2
n= 2
= 2
= 219.19
e 5
2. A production line operation is designed to fill cartons with laundry detergent to a mean
weight of 32 ounces. A sample of cartons is periodically selected and weighed to
determine whether underfilling or overfilling is occurring. If the sample data lead to a
conclusion of underfilling or overfilling, the production line will be shut down and adjusted
to obtain proper filling. Formulate the null and alternative hypotheses that will help in
deciding whether to shut down and adjust the production line.
Solution: H0: μ = 32 & Ha: μ ≠ 32
One sample Z-test: Example-1
= 0.05/2 = 0.05/2
X−μ
ZSTAT =
σ
n
2. Dr Zeppo grades his introductory statistics class on a curve. Let’s suppose that the average grade in
his class is 67.5, and the standard deviation is 9.5. Of his many hundreds of students, it turns out
that 30 of them also take psychology classes. Do the psychology students tend to get the same
grades as everyone else?
3. In order to determine whether two varieties of pomegranates significantly differ in their mean
weights, 70 random samples were drawn from each variety and their mean weights measured. The
mean weight of variety-A was measured to be 220.6 gram with standard deviation of 24.6 grams,
while the mean weight of variety-B was 232.8 gram with a standard deviation of 27.8 gram. Test
whether the mean weights are different to a significance level of 0.05.