U 2 Written Assignment Unit 2
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University of the People *
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CPH 4510
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Health Science
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Dec 6, 2023
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HS 4510-01 Biostatistics - AY2024-T2
Written Assignment Unit 2
1.
What is the proportion of males and females in this sample?
Because the total number of males in the study is 11 and the total number of females
is 10, the proportion of males in this sample is 11/25 (0.44) and the proportion of
females in this sample is 10/25 = 0.4.
2.
Suggest two different graphs that can be used to display the data in Question 1.
Explain your answer.
a.
A histogram is a useful graph for showing relative frequencies and the
average pulse rate because it is particularly effective for displaying data that is
ranked or ordered (Taylor 2019).
b.
A bar chart can be used to show the sex of the participants because it is a
suitable graphical representation for categorical data. According to Taylor (2019), a
bar graph depicts the frequency of categorical data using bars.
3.
a.
Mean of males – 79+90+75+60+65+79+54+98+49 = 649/9 = 72.1
Mean of females – 66+59+79+95+72+55+64+60+65=615/9 = 68.3
Mean of both male and female
= 49+54+55+59+60+60+64+65+65+66+72+75+79+79+79+90+95+98 = 1264
= 1264/18 = 70.2
b.
Male pulse
49, 54, 60, 65, 75, 79, 79, 90, 98
Male median= 75
Female pulse
55, 59, 60, 64, 65, 66, 72, 79, 95
Female Median= 65
Pulse for female and male
49,54,55,59,60,60,64,65,65,66,72,75,79,79,79,90,95,98
Median for female and male
= 65+66 = 131/2 = 65.5
c.
Male pules 49, 54, 60, 65, 75, 79, 79, 90, 98
Male median = 75
Female Pules 55, 59, 60, 64, 65, 66, 72, 79, 95
Female median = 65
Male and female Pules 49,54,55,59,60,60,64,65,65,66,72,75,79,79,79,90,95,98
Male and female median = 65+66 = 131/2 = 65.5
c.
Male degree of freedom
= n – 1 = 9-1 = 8
Female degree of freedom = n – 1 = 9 – 1 = 8
Degrees of freedom of both male and female = 9 + 9 = 18, 18 – 1 = 17
d.
Female =X , x = {55, 59, 60, 64, 65, 66, 72, 79, 95}
= 55+59+60+64+65+66+72+79+95 = 615 / 9 = 68.3
= (55-68.3)
2
+ (59-68.3)
2
+ (60-68.3)
2
+ (64-68.3)
2
+ (65-68.3)
2
+ (66-68.3)
2
+ (72-
68.3)
2
+ (79-68.3)
2
+ (95-68.3)
2
= 176.89 + 86.49 + 68.89 + 18.49 + 10.89 + 5.29 + 13.69 + 114.49 + 712.89
= 1208.01/9-1 =1208.01/8 = 151
= sqrt 151 = 12.29
SD of females = 12.29
Male=y,
y= {49, 54, 60, 65, 75, 79, 79, 90, 98}
= 49+54+60+65+75+79+79+90+98 = 649/9 = 72.1
= (49-72.1)
2
+ (54-72.1)
2
+ (60-72.1)
2
+ (65-72.1)
2
+ (75-72.1)
2
+ (79-72.1)
2
+ (79-
72.1)
2
+ (90 -72.1)
2
+ (98-72.1)
2
= 533.61 + 327.61 + 146.41 +50.41 + 8.41 + 47.61 + 47.61 + 320.41 + 670.81
=2152.89/9-1 = 2152.89/8 = 269.11
= sqrt 269.11 = 16.4
SD of males = 16.4
e.
Because all of the values in the distribution were used, the variance and standard
deviation that were calculated offered us precise information that could be
depended on, the measure of variability is a continuous variable.
f.
Range for male and female is 98-49=49
Interquartile range (IQR) for both male and female is 49,54,55,59|,60,60,64,65,65,|
66,72,75,79,79,|79,90,95,98
Q1 location = (n+1) / 4 = (18+1)/4 = 4.75, this falls between 59 and 60,
Q3 location = 3(n+1)/4 = 3(18+1)/4 = 14.25, falls between 79 and 79.
Estimating Q1 = 59+3/4(60-59) = 59.75
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Estimating Q2 = 79+1/3(79-79) = 79
I.Q.R = Q3 – Q1 = 79 – 59.75 = 19.25
4.
a.
Intensive care and surgery mean length of stay –
(26*10)+(12*20)/30 = 500/30 = 16.7 Intensive care
and surgery patients spend an average of 16.7 days in
the hospital.
b.
Mean length of everything - [(26*10) + (12*20) + (4*40) + (8*50)]/120 = 1060/120
= 8.83
Mean length of everything is = 8.83
Reference
Taylor, C. (2019). Pie Charts, Histograms, and Other Graphs Used in Statistics.
Retrieved from https://www.thoughtco.com/frequently-used-statistics-graphs-
4158380