SOLUTION: Considering the following distribution for blood pressure level of male patients in a given hospital (in mm Hg). Values (BP) Frequency (# of patients) 140- 150 17 150- 160 29

Question 1201046: Considering the following distribution for blood pressure level of male patients in a
given hospital (in mm Hg).
Values (BP) Frequency (# of patients)
140- 150 17
150- 160 29
160- 170 42
170- 180 72
180- 190 84
190- 200 107
200- 210 49
210- 220 34
220- 230 31
230- 240 16
240- 250 12
Calculate:
a) Calculate the average blood pressure of male patients in a given hospital.
b) All the quartiles.
c) Find the 8th deciles.
d) Find the 20th percentiles.
Answer by asad1975(2) (Show Source): You can put this solution on YOUR website!
To answer these questions, I will use the formula for finding the mean, quartiles, deciles and percentiles of grouped data. The formula is:
$$x_p = L + \frac{pN - F}{f} \times W$$
where $x_p$ is the value of the pth percentile, L is the lower boundary of the class interval containing the pth percentile, p is the percentile expressed as a decimal (e.g. 0.25 for 25th percentile), N is the total number of observations, F is the cumulative frequency of the class interval before the one containing the pth percentile, f is the frequency of the class interval containing the pth percentile, and W is the width of the class interval.
To use this formula, I will first construct a frequency table with the cumulative frequencies and class boundaries for each class interval.
Values (BP) Frequency (# of patients) Cumulative Frequency Class Boundaries
140- 150 17 17 139.5 - 150.5
150- 160 29 46 150.5 - 160.5
160- 170 42 88 160.5 - 170.5
170- 180 72 160 170.5 - 180.5
180- 190 84 244 180.5 - 190.5
190- 200 107 351 190.5 - 200.5
200- 210 49 400 200.5 - 210.5
210- 220 34 434 210.5 - 220.5
220- 230 31 465 220.5 -230.5
230-240 16 481 230.5 -240.5
240-250 12 493 240.5 -250.5
The total number of observations is N =493.
a) To calculate the average blood pressure of male patients in a given hospital, I will use the mean formula for grouped data, which is:
$$\bar{x} = \frac{\sum f x}{N}$$
where $\bar{x}$ is the mean, f is the frequency of each class interval, x is the midpoint of each class interval, and N is the total number of observations.
To find the midpoint of each class interval, I will add the lower and upper boundaries and divide by two.
Values (BP) Frequency (# of patients) Midpoint (x)
140-150 17 145
150-160 29 155
160-170 42 165
170-180 72 175
180-190 84 185
190-200 107 195
200-210 49 205
210-220 34 215
220-230 31 225
230-240 16 235
240-250 12 245
To find the product of frequency and midpoint for each class interval, I will multiply f and x.
| Values (BP) Frequency (# of patients) Midpoint (x) f x
140-150 17 145 2465
150-160 29 155 4495
160-170 42 165 6930
170-180 72 175 12600
180-190 84 185 15540
190-200 107 195 20865
200-210 49 205 10045
210-220 34 215 7310
220-230 31 225 6975
230-240 16 235 3760
240-250 12 245 2940
The sum of f x is $\sum f x = $ 89425.
To find the mean, I will plug in the values into the formula:
$$\bar{x} = \frac{\sum f x}{N}$$
$$\bar{x} = \frac{89425}{493}$$
$$\bar{x} = 181.41$$
The average blood pressure of male patients in a given hospital is 181.41 mm Hg.
b) To find the quartiles, I will use the formula for finding the percentiles with p = 0.25, 0.5, and 0.75.
For the first quartile (Q1), p = 0.25. I will plug in the values into the formula:
$$x_p = L + \frac{pN - F}{f} \times W$$
$$x_{0.25} = L + \frac{0.25 \times 493 - F}{f} \times 10$$
The class interval containing the 25th percentile is 170-180, since its cumulative frequency is the first one greater than or equal to 0.25 x 493 = 123.25.
The lower boundary of this class interval is L = 170.5.
The cumulative frequency of the class interval before this one is F = 88.
The frequency of this class interval is f = 72.
The width of this class interval is W = 10.
Plugging in these values, I get:
$$x_{0.25} = 170.5 + \frac{0.25 \times 493 - 88}{72} \times 10$$
$$x_{0.25} = 170.5 + \frac{35.25}{72} \times 10$$
$$x_{0.25} = 170.5 + 4.9$$
$$x_{0.25} = 175.4$$
The first quartile (Q1) is 175.4 mm Hg.
For the second quartile (Q2), p = 0.5. I will plug in the values into the formula:
$$x_p = L + \frac{pN - F}{f} \times W$$
$$x_{0.5} = L + \frac{0.5 \times 493 - F}{f} \times 10$$
The class interval containing the 50th percentile is 180-190, since its cumulative frequency is the first one greater than or equal to 0.5 x 493 = 246.5.
The lower boundary of this class interval is L = 180.5.
The cumulative frequency of the class interval before this one is F =160.
The frequency of this class interval is f =84.
The width of this class interval is W =10.
Plugging in these values, I get:
$$x_{0.5} =180.5 + \frac{0.5 \times493 -160}{84} \times10$$
$$x_{0.5} =180.5 + \frac{86.5}{84} \times10$$
$$x_{0.5} =180.5 +10.3$$
$$x_{0.5} =190.8$$
The second quartile (Q2) is 190.8 mm Hg.
For the third quartile (Q3), p = 0.75. I will plug in the values into the formula:
$$x_p = L + \frac{pN - F}{f} \times W$$
$$x_{0.75} = L + \frac{0.75 \times493 - F}{f} \times10$$
The class interval containing the 75th percentile is200-210, since its cumulative frequency is the first one greater than or equal to 0.75 x493=369.75.
The lower boundary of this class interval is L=200.5.
The cumulative frequency of the class interval before this one is F=351.
The frequency of this class interval is f=49.
The width of this class interval is W=10.
Plugging in these values, I get:
$$x_{0.75}=200 .5+ \frac{0 .75\times493-351}{49}\times10 $$
$$x_{0 .75}=200 .5+ \frac{18 .75}{49}\times10 $$
$$x_{0 .75}=200 .5+3 .83 $$
$$x_{0 .75}=204 .33 $$
The third quartile (Q3) is 204 .33 mm Hg.
c) To find the eighth decile (D8), p=0 .8 . I will plug in the values into the formula:
$$x_p=L+\frac{pN-F}{f}\times W $$
$$x_{0 .8}=L+\frac{0 .8\times493-F}{f}\times10 $$
The class interval containing the80th percentile is200-210, since its cumulative frequency is the first one greater than or equal to 0

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