Chapter 1.3, Problem 97E

(a)

To determine

To find: the standard deviation

Answer to Problem 97E

The standard deviation is $0.6419mg/dl$

Explanation of Solution

Given:

Measurements of the level of phosphate in the blood, made on 6 consecutive visits to a clinic:5.6, 5.2, 4.6, 4.9, 5.7, 6.4.

Calculation:

The sample size is the number of values: $n=6$

The mean is the sum of all values divided by the sample size: $\overline{x}=\frac{5.6+5.2+4.6+4.9+5.7+6.4}{6}=5.4mg/dl$

If $x_i$ are the data values, then $x_i-\overline{x}$ are the deviation ${\left(x_i-\overline{x}\right)}^2$ the squared deviation:

$x_i$	$\left(x_i-x_{mean}\right)$	${\left(x_i-x_{mean}\right)}^2$
5.6	0.2	0.04
5.2	-0.2	0.04
4.6	-0.8	0.64
4.9	-0.5	0.25
5.7	0.3	0.09
6.4	1	1

The variance is the sum of the squared deviations divided by $n-1$ :

  $s_x{}^2=\frac{0.04+0.04+0.64+0.25+0.09+1}{6-1}=\frac{2.06}{5}=0.412$

The standard deviation is the square root of the variance:

  $s_x=\sqrt{0.412}\approx 0.6419mg/dl$

Conclusion:

Hence,the standard deviation is $0.6419mg/dl$

(b)

To determine

To interpret:the value of $s_x$ .

Answer to Problem 97E

Phosphate level varies from the mean by .6419 mg/dl.

Explanation of Solution

Calculation:

The phosphate level typically varies from the mean by 0.6419 mg/dl.