Question

The National Football League (NFL) polls fans to develop a rating for each football game. Each game is rated on a scale from 0 (forgettable) to 100 (memorable). The fan ratings for a random sample of 12 games follow. 57 61 87 74 72 73 19 56 81 79 83 75 a. Develop a point estimate of mean fan rating for the population of NFL games (to 2 decimals). b. Develop a point estimate of the standard deviation for the population of NFL games (to 4 decimals).

Answer 1

Answer:

a:  68.08

b:  5.3065

Step-by-step explanation:

The point estimate is a single value.  The point estimate for the mean is the mean of the sample.  In this case, the mean is the sum of the values divided by the total number of values.  The point estimate is the standard deviation of the sample.  The standard deviation is the square root of the variance.  

See the attached photo for the calculations of these values.

Answer 2

Answer:

a. The estimate of  mean  fan rating for the population of NFL games=68.08

b. The estimate of standard deviation=18.8785

Step-by-step explanation:

Given

The fan ratings for a random sample of 12 games follow:

57, 61, 87,74,72,73,19,56,81,79,83 and 75

a.Mean =$\frac{ sum\;of\;data}{ total\;number \; of data}$

Mean=$\frac{57+61+87+74+72+73+19+56+81+79+83+75}{12}$

Mean= 68.08

b.$\mid x-\bar x\mid$                                    ${\mid x-\bar x\mid}^2$

11.08                                                122.7664

7.08                                                 50.1264

18.92                                                357.9664

5.92                                                 35.0464

3.92                                                  15.3664

4.92                                                  72.6192

49.08                                                2408.8464

12.08                                                 145.9264

12.92                                                  166.9264

10.92                                                  119.2464

14.92                                                   734.064

6.92                                                    47.8864

Standard deviation=${\sqrt\frac{\sum{\mid x-\bar x\mid}^2}{n}$

n=12

$\sum{\mid x-\bar x\mid}^2=4276.7872$

Standard deviation=$\sqrt\frac{4276.7872}{12}$

Standard deviation=$\sqrt{356.3989}$

Standard deviation = 18.8785

The estimate of the standard deviation for the population of NFL games=18.8785