Answer to Question #210661 in Accounting for Olyver Aseri

Question

1a)The following distribution represents the number of minutes spent by a herd of cows in grazing

Minutes/week 0-99 100-199 200-299 300-399 400-499 500-599 600 & more

Number of cows 21 26 59 72 52 26 8

Determine the median.

b)If coveriance between X and Y is 18.8 and the variance of X and Y are 22.4 and 26.8 respectively. Find the coefficient of correlation between them.

2a) Discuss the importance coefficient of correlation in explaining the relationship between two variables.

b)The chance of any photocopier being defective is 20%. If 15 photocopiers are selected at random, what is the probability that

I)Fewer than 4 will be defective?

ii)None will be defective?

iii) Exactly 6 will be defective?

iv) Between 7 and 9 will be defective?

v)At least 10 will not be defective?

C) Determine the probability of receiving a score greater than 850 on GMAT test that has a mean of 496 and the standard deviation of 116

Answer 1

Solution:

1.a). The median value:

 

Median = $L + (\frac{\frac{1}{2}N - F }{fm})\times C$

Where: L = Lower boundary

           N = Total frequencies

           F = Total frequency above box

           fm = Frequency in box

           C = Class interval size

$=\frac{1}{2}\times 264 = 132$

132 falls under CF of 178

 

L = $\frac{299 + 300 }{2} = 299.5$

N = 264

F = 106

fm = 72

C = 399 – 299 = 100

Median = $299.5 + (\frac{\frac{1}{2}\times264 - 106 }{72})\times 100 = 299.5 + 36.1 = 335.6$

 

Median value = 335.6

b.). Coefficient correlation (X, Y):

$r_{X,Y} = \frac{S_{XY} }{(S_{X}S_{Y}) }$

Where:

·        r_XY = Correlation between X and Y

·        S_XY = Covariance between X and Y

·        S_X = Standard deviation of X

·        S_Y = Standard deviation of Y

Covariance between X and Y (S_XY) = 18.8

Standard deviation = Square root of variance

Variance of X = 22.4

Variance of Y = 26.8

Standard deviation of X = $\sqrt{22.4} = 4.7329$

Standard deviation of Y = $\sqrt{26.8} = 5.1769$

Coefficient correlation (r,_XY) = $\frac{18.8 }{(4.7329)(5.1769) } = \frac{18.8 }{24.5014 } = 0.7673$

Coefficient of correlation between X and Y = 0.7673

2.a.). The coefficient of correlation is very important in measuring and analyzing the strength of the relationship between two variables, including the direction of the linear relationships between variables pairs. The coefficient of correlation is positive when one variable increases as the other increases, that is when the two variables are moving in the same direction. On the other hand, the coefficient of correlation is negative when one variable increases and the other variable decreases, which is when the two variables are moving in opposite directions.

b.). i). Fewer than 4 will be defective:

This means the probability that 3 and below will be found defective:

The probability of a photocopier being defective = 20$\%$ = 0.2

The probability of a photocopier being good or not defective = 80$\%$ = 0.8 that is, (1 – 0.2)

N = 15

Lex X be the number of defective items.

Let the probability of a photocopier being defective P(X)

r = number of items being chosen

P (X ≤ 4) = $(\frac{n!}{r!\times (n - r)!} ) \times0.20^{3} \times0.80^{12}$

P (X ≤ 4) = $(\frac{15!}{3!\times (15 - 3)!} ) \times0.20^{3} \times0.80^{12}$ = $(\frac{15!}{3!\times 12!} ) \times0.20^{3} \times0.80^{12} = 0.2501$

The probability that fewer than 4 will be defective = 0.2501

ii). None will be defective:

P (X =0) = $(\frac{15!}{0!\times 15!} ) \times0.20^{0} \times0.80^{15} = 0.80^{15} = 0.0352$

The probability that none will be defective = 0.0352

iii.). Exactly 6 will be defective:

P (X = 6) = $(\frac{15!}{6!\times 9!} ) \times0.20^{6} \times0.80^{9} = 0.0430$

The probability that exactly 6 will be defective = 0.0430

iv.). Between 7 and 9 will be defective:

P (X = 7) + P (X = 8) + P (X = 9) = $((\frac{15!}{7!\times 8!} ) \times0.20^{7} \times0.80^{8}) + ((\frac{15!}{8!\times 7!} ) \times0.20^{8} \times0.80^{7}) + ((\frac{15!}{9!\times 6!} ) \times0.20^{9} \times0.80^{6}))$ =0.0138 + 0.0035 + 0.00067 = 0.011797

The probability that between 6 and 7 will be defective = 0.011797

v.). At least 10 will be defective:

P (X ≥ 10) = 1 – P (X = 10)

=$1 - (\frac{15!}{10!\times 5!} ) \times0.20^{10} \times0.80^{5} = 1 - 0.00010 = 0.9999$

The probability that at least 10 will be defective = 0.9999

C.). Probability = $\frac{(X - Mean)}{Standard\; deviation}$

X = 850

Mean = 496

Standard deviation = 116

Probability = $\frac{(850 - 496)}{116} = \frac{354}{116} = 3.05$