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Sta301 assignment 1 solution spring 2021 pdf.

Question 1: By using the following frequency distribution of weights distribution of 120 students at Virtual University find. 1        Mean Deviation from Mean 2.       Co-efficient of Mean Deviation from Means Weights (Kg) Frequency 45-49 1 50-54 4 55-59 17 60-64 28 65-69 25 70-74 18 75-79 13 80-84 6 85-89 5 90-94 2 95-99 1 Read Online   👇👇👇👇👇

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STA301 Assignment No 1 Solution Fall 2023

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Sta301 assignment no 1 solution fall 2023 – vu assignment solution – vu fall assignment solution.

The STA301 Assignment No 1 Solution Fall 2023  of Virtual University (VU) course assignment, and I will share with you today the solution I came up with. Always come back to StudySolution for the most recent updates regarding the answers to your assignments.

STA301 Assignment No 1 Solution Fall 2023

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Assignment No. 1

Solution Fall 2023

STA301 Assignment 1

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Students Can Get STA301 Assignment No 1 Solution Fall 2023 just By Visiting below.

Get Your Solution Below

Question no 1:

Construct a stem-and-leaf display for the grades in Statistics of the students in a college.

96 93 88 117 127 95 113 96 108 139 142
94 107 125 155 155 103 112 112 135 132 111
125 104 106 139 134 118 136 125 143 120 103
113 124 138 94 127 119 148 117 97 156 120
9 6   5   3   4   4   6   7
10 8   7   3   4   6   3
11 7  3  2  2  1  8  3  9  7
12 7  5  5  5  4  0  0   7
13 9  5  3  8   4  6  8
14 2    3    8

Question no 2:

The following table is constructed from the weights (recorded to the nearest pound) of 40 male students of a college. Calculate the mean and median of the distribution.

118-126 3
127-135 5
136-144 9
145-153 12
154-162 5
163-171 4
172-180 2

Calculate the mean and median

118-126 122 3 366
127-135             131 5 655
136-144             140 9 1260
145-153             149 12 1788
154-162             158 5 790
163-171  167 4 668
172-180             176 2 352
    = 40 =  5879

Now find Median

L = lower class boundary of the median class

h = class interval

f = frequency of the median class

n = total number of observation

c = cumulative frequency of the class preceding the median class

118-126 117.5-126.5 3 3
127-135 126.5-135.5 5 8
136-144 135.5-144.5 9 17
145-153 144.5-153.5 12 29
154-162 153.5-162.5 5 34
163-171 162.5-171.5 4 38
172-180 171.5-180.5 2 40

Now, putting the value in formula

= 144.5+2.25

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