Statistics: Solved Assignments and Essays

This document contains solutions to three Statistics assignments. The first assignment involves calculating the probability of a wrong order being taken by three different ladies. The second assignment involves proving a formula for calculating the expected value of a discrete random variable. The third assignment involves calculating the sample correlation coefficient between two variables and justifying the need for financial assistance to a student based on their correlation coefficient.

Statistics
Name:
Institution:
20th February 2019

Assignment
1. Three ladies (W,X,Y) take special orders at the college bookstore. W handles 50% of
special orders filed and makes a mistake with probability 0,01. The corresponding
numbers for X and Y are 0.2 , 0.3 and 0.04 , 0.02 respectively. If a wrong order has been
taken what is the probability it was handled by W, X, Y?
Answer
P ( W ) =0.01∗0.5=0.005
P ( X ) =0.04∗0.2=0.008
P ( Y )=0.02∗0.3=0.006
2. Show that E(ax+b)=aE(x) +b, where X is a discrete r.v with Pmf f(x) and a, b constants.
Toss a coin 3 times and define X such that X=number of “Heads” that occur in this
experiment. Find E(X), var(X). Sample space 8.
Answer
We need to prove that;
E ( ax +b )=aE ( x ) +b
E ( ax +b )=∑
x
( ax +b ) p(x )
¿ ∑
x
( ax . p ( x ) +b . p(x ) )
¿ ∑
x
ax . p (x)+∑
x
b . p(x )
¿ a ( ∑
x
x . p ( x) ) + b ( ∑
x
p(x) )
We know that
(∑
x
x . p( x )
)=E( x)
( ∑
x
p( x) ) =1
Therefore we have;
¿ a (∑
x
x . p ( x) )+ b (∑
x
p(x) )=a E ( x ) +b
Hence we have proved that;
E ( ax +b ) =a E ( x ) +b

Tossing a coin
Possible outcomes;
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
P ( X=0 )= 1
8
P ( X=1 )=3
8
P ( X=2 ) = 3
8
P ( X=3 )= 1
8
E ( X ) =( 0∗1
8 )+( 1∗3
8 )+( 2∗3
8 )+ (3∗1
8 )=12
8 =1.5
Var ( X )= ( 02∗1
8 )+ ( 12∗3
8 )+ ( 22∗3
8 )+ ( 32∗1
8 )−1.52
Var ( X )= 24
8 −2.25=3−2.25=0.75
3.
Xi (2,4,6,8,10,12,14,16,18,20)
Yi (10,20,50,40,50,60,80,90,90,120)
a. Calculate the sample correlation coefficient between Y (quantity supplied) and X
(price) from the above sample information.
Answer
r = n ( ∑ xy ) −∑ x ∑ y
√ [ n∑ x2− ( ∑ x )
2
] [ n∑ y2− ( ∑ y ) 2
]
∑ xy =8520 ,∑ x=110 ,∑ y=610 ,∑ x2=1540 ,∑ y2=47700
r = n (∑ xy )−∑ x ∑ y
√ [ n∑ x2− (∑ x )2
] [ n∑ y2− (∑ y )2
] = 10∗8520−110∗610
√ [10∗1540−1102 ] [ 10∗47700−6102 ] = 18100
18605.644 =0.9
b. By gathering information on the performance of a student in various exams and
the number of hours he worked during the academic year, we compute the
correlation coefficient which assumes the value of -0.9. Does this finding justify

giving financial assistance to this student in order to improve his grade
performance? Explain yourself.
Answer
The results do justify giving financial assistance to this student in order to
improve his grade performance. This is based on the fact that an increase in the
time spent by the student working leads to poor performance in exams by the
student. This probably means that the student spends more time looking for
money and so he/she has less time to read hence the poor performance. So this
justifies that giving the student financial assistance would probably help the
student have more time to read for exams and hence perform better.