Question

Five pulse rates are randomly selected from a set of measurements. The five pulse rates have a mean of 72.472.4 beats per minute. Four of the pulse rates are 7878​, 6060​, 9191​, and 8282. a. Find the missing value. b. Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected. How many of the n values can be freely assigned before the remaining values are​ determined? (The result is referred to as the number of degrees of​ freedom.)

Answer 1

The missing pulse rate is 51 beats per minute. When creating a list of 'n' values with a known mean, you can freely assign (n - 1) values before the remaining value is determined, which is referred to as the number of degrees of freedom.

Step-by-step explanation:

To find the missing pulse rate, we use the formula for the mean of a set of numbers. The mean (μ) is the sum of the values in the set divided by the number of values in the set (n). Given that the mean of the five pulse rates is 72.4 and we already know four of them, we can set up an equation to find the missing value (x):

(78 + 60 + 91 + 82 + x) / 5 = 72.4

Multiplying both sides by 5 gives us:

78 + 60 + 91 + 82 + x = 362

Now, we add up the known values:

78 + 60 + 91 + 82 = 311

We subtract this sum from 362 to find the missing value:

x = 362 - 311 = 51

Therefore, the missing pulse rate is 51 beats per minute.

Regarding the number of freely assignable values, the number of freely assignable values is (n - 1), where n is the total number of values in your data set. This is because once you have (n - 1) values, the last value is determined by the requirement that the mean of the set is a specific value. This concept is known as the degrees of freedom.

Answer 2

a) 330951

b) n-1

Step-by-step explanation:

a)

`\bar{X} = 72472,4 \\ x_1 = 7878, x_2 = 6060,x_3 = 9191 , x_4 = 8282, x_5 = ?`

`\bar{X} = (1)/(n) \sum_(i=1)^n x_i \\72472,4 = (1)/(5) (7878 + 6060 + 9191 + 8282 + x_5) \\362362 = 31411 + x_5 \\x_5 = 330951`

b)

Imagine the following situation:

`n = 5 \\ \bar{X} = 25 \\ (1)/(5)\cdot (13+a+b+c+d) = 25`

We realize that we can assign a maximum of 4 values ​​so that the average is 25. That is, when we set

one value, we have n-1 possibilities to assign other values.