Statistics Assignment: Probability, Random Variables, and Correlation

This statistics assignment addresses three distinct problems. The first calculates the probability of a wrong order being handled by one of three ladies (W, X, Y) at a college bookstore, given their individual error rates and order volumes. The second proves the linearity of expectation, E(ax+b)=aE(x)+b, for a discrete random variable and then calculates the expected value and variance for the number of heads obtained from tossing a coin three times. The third problem involves calculating the sample correlation coefficient between quantity supplied (Y) and price (X) using provided data, and discusses the implications of a negative correlation between a student's exam performance and working hours, suggesting financial assistance to improve academic performance.

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.