contestada

Six different​ second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic readings​ (in mmHg) are listed below. Find the​ range, variance, and standard deviation for the given sample data. If the​ subject's blood pressure remains constant and the medical students correctly apply the same measurement​ technique, what should be the value of the standard​ deviation? 127 150 121 120 140 128

Respuesta :

lucic

Answer:

1. Range =30

2. Variance =137.6

3. Standard deviation=11.7303

Step-by-step explanation:

This question requires you to find the range, variance and standard deviation of sample data set.

Given the data as; 127 150 121 120 140 128

Arrange the data in ascending order;

sample set S={120, 121, 127, 128, 140, 150}

number of elements, n=6

1. Range = Maximum (S) - Minimum (S) = 150- 120 = 30

⇒Find the mean of the data set

[tex]mean= \frac{120+121+127+128+140+150}{6} = 786/6 = 131[/tex]

2. Variance is the measure of how far a set of data is spread out.Standard deviation is the square-root of variance.To find variance you need to follow the steps below;

  • Find the mean of the sample data
  • Find the deviation of each of the data from the mean
  • Square each value of the deviations from the mean
  • Find the sum in the values of the squared deviations
  • Divide the sum in the values of the squared deviations by n-1 where n is the number of elements to get the variance
  • Find the square-root of the variance to get the standard deviation of the sample data

Finding the deviations from the mean and their squares

Deviations             Squares of deviations

120-131= -11               -11²=   121

121-131= -10                -10² =100

127-131= -4                  -4² = 16

128-131= -3                   -3= 9

140-131= 9                    9²=  81

150-131= 19                19²=  361

Finding the sum of the squares of the deviations from the mean

[tex]=121+100+16+9+81+361=688[/tex]

Finding the variance

Variance, S²=(sum of squares of deviations from mean)/ n-1

[tex]=\frac{688}{n-1} =\frac{688}{6-1} =\frac{688}{5} =137.6[/tex]

Finding standard deviation

Standard deviation , s , is the square-root of the variance 

[tex]s=\sqrt{137.6} =11.73[/tex]

Otras preguntas