Answer to Question #174271 in Statistics and Probability for Kwame Hay

Question

Question 2

You are required to use the data provided in Table 1.0 to determine:

a) The modal price in $US AN( 7 marks)

b) The standard deviation AN( 7 marks)

c) The skewness of frequency distribution and explain the significance of the results in terms the direction of your findings. AN (6 marks)

Explain the significance of your answer in each case.

Table 1.0

Price ($US)

Frequency

1.0-1.01

1.05-1.09

1.10-1.14

1.15-1.19

1.20-1.24

1.25-1.29

4 6 1 0 15 8 5

Total Marks: 20

Answer 1

a) the modal group (the group with the highest frequency): $1.15-1.19$

Estimated Mode (modal price)$=L+\frac{f_m-f_{m-1}}{(f_m-f_{m-1})+(f_m-f_{m+1})}\cdot w$

where

$L$ is the lower class boundary of the modal group

$f_{m-1}$ is the frequency of the group before the modal group

$f_{m}$ is the frequency of the modal group

$f_{m+1}$ is the frequency of the group after the modal group

$w$ is the group width

Price that occurs most often:

modal price$=1.145+\frac{15-1}{15-1+15-8}\cdot0.05=1.178$

b) Grouped standard deviation:

$\sigma=\sqrt{\frac{\sum FM^2-n\mu^2}{n-1}}$

$F$ is frequency

$M$ is midpoint of range

$\mu$ is data mean

$M=\frac{x_1+x_2}{2}$

$M_1=1.005,M_2=1.03,M_3=1.07,M_4=1.12,M_5=1.17,M_6=1.22,M_7=1.27$

$\mu=\sum MF/n$

$n=\sum F=4+6+1+15+8+5=39$

$\mu=\frac{1.005\cdot4+1.03\cdot6+1.07+1.17\cdot15+1.22\cdot8+1.27\cdot5}{39}=1.15$

$\sum FM^2=1.005^2\cdot4+1.03^2\cdot6+1.07^2+1.17^2\cdot15+1.22^2\cdot8+1.27^2\cdot5=52.06$

Standard deviation (dispersion of a set of prices):

$\sigma=\sqrt{\frac{52.06-39\cdot1.15^2}{38}}=0.112$

c) Skewness, in statistics, is the degree of asymmetry observed in a probability distribution.

$Skew=3(Mean-Median)/Standard\ Deviation$

Estimated Median=$L+\frac{(n/2)-B}{G}\cdot w$

$B$ is the cumulative frequency of the groups before the median group

$G$ is the frequency of the median group

the median group: $1.15-1.19$

$B=4+6+1=11$

Estimated Median$=1.145+\frac{(39/2)-11}{15}\cdot0.05=1.173$

$Skew=3(1.15-1.173)/0.112=-0.625$

So, the distribution left side skewed.