Answer to Question #282111 in Statistics and Probability for Vimal

Question

3.2 4.7 3.8 4.3 3.5 5.0 4.3 5.2 5.1 4.2

1.Constuct a frequency destribution mode.

2.Determined the interval where k=10.

3.Compute for mean,median ,mode ,variance and standard deviation of grouped date

Answer 1

3.2 4.7 3.8 4.3 3.5 5.0 4.3 5.2 5.1 4.2 1.8 5.1 5.8 5.0 5.6 4.5 4.8 4.0 2.7 3.8 4.2 4.1 2.7 5.6 6.0 3.9 4.2 3.4 5.5 3.8 5.1 4.6 5.8 4.3 2.5 5.8 4.5 4.9 4.7 4.4 2.4 3.4 4.9 4.3 3.7 4.9 2.7 3.0 4.2 4.0 3.6 3.9 3.1 3.9 5.3 4.4 3.7 3.1 2.8 3.8

Least to Greatest Value:

1.8, 2.4, 2.5, 2.7, 2.7, 2.7, 2.8, 3, 3.1, 3.1, 3.2, 3.4, 3.4, 3.5, 3.6, 3.7, 3.7, 3.8, 3.8, 3.8, 3.8, 3.9, 3.9, 3.9, 4, 4, 4.1, 4.2, 4.2, 4.2, 4.2, 4.3, 4.3, 4.3, 4.3, 4.4, 4.4, 4.5, 4.5, 4.6, 4.7, 4.7, 4.8, 4.9, 4.9, 4.9, 5, 5, 5.1, 5.1, 5.1, 5.2, 5.3, 5.5, 5.6, 5.6, 5.8, 5.8, 5.8, 6

1.

2.

interval where cumulative frequency k=10:

class 3 - 4

3.

mean:

$\mu=\sum x_if_i=4.22$

where x_i is midpoint of class,

f_i is frequency

for Median Class:

value of (n/2)th observation = value of (60/2)th observation =

= value of 30th observation

The median class is 4-5

median:

$m=L+\frac{n/2-cf}{f}c=4+\frac{30-24}{22}=4.27$

where

L=lower boundary point of median class =4

n=Total frequency =60

cf=Cumulative frequency of the class preceding the median class =24

f=Frequency of the median class =22

c=class length of median class =1

for Mode Class:

maximum frequency is 22.

The mode class is 4-5

mode:

$M=L+\frac{f_1-f_0}{2f_1-f_0-f_2}c=4+\frac{22-17}{2\cdot22-17-13}=4.36$

Variance:

$\sigma^2=\frac{\sum x^2_if_i-\sum x_if_i/n}{n}=1.04$

standard deviation:

$\sigma=\sqrt{\frac{\sum x^2_if_i-\sum x_if_i/n}{n}}=1.018$