Question

For a population, the mean is 19.4 and the standard deviation is 5.8. Compare the mean and standard deviation of the following random samples to the population parameters.

12, 15, 17, 20, 13, 11, 18, 19, 15, 14

Answer 1

1. 25, 32, 16, 12, 11, 38, 22, 21, 19, 20 mean: 21.6, standard deviation: 8.4 The mean is higher than the population and the standard deviation is greater, which means that the data values in the sample are more spread out than the population and 3. 4, 5, 3, 11, 24, 19, 8, 7, 11, 10 mean: 10.2, standard deviation: 6.68 The mean is lower and the standard deviation is a little bit greater, which means that the data values in the sample are more spread out than the population. For a population, the mean is 56.6 and the standard deviation is 8.5. Compare the mean and standard deviation of the following random samples to the population parameters. 5. 42, 48, 52, 56, 69, 74, 50, 49 mean: 55, standard deviation: 10.99 The mean is lower and the standard deviation is higher, which means that the data values are sample are more spread out than the population so tell me if i helped

Answer 2

The mean and standard deviation of random samples to the population parameters is less than the mean and standard deviation of population.

Explanation:

Given : For a population, the mean is 19.4 and the standard deviation is 5.8.

The following random samples to the population parameters.

12, 15, 17, 20, 13, 11, 18, 19, 15, 14

To find : Compare the mean and standard deviation ?

Solution :

1) We find the mean of the random sample,

`M=(\sum x_n)/(n)`

`M=(12+15+17+20+13+11+18+19+15+14)/(10)`

`M=(154)/(10)`

`M=15.4`

The Mean of random sample 15.4 is less than 19.4 the mean of population.

2) We find the standard deviation of the random sample,

`s=\sqrt{(1)/(n-1)\sum_(i=1)^(n)(x-\overline{x})^(2)}`

`s=\sqrt{(1)/(10-1)\sum_(i=1)^(n)(x-\overline{x})^(2)}`

Solving summation separately,

`\sum_(i=1)^(n)(x-\overline{x})^(2)=(12-15.4)^2+(15-15.4)^2+(17-15.4)^2+(20-15.4)^2+(13-15.4)^2+(11-15.4)^2+(18-15.4)^2+(19-15.4)^2+(15-15.4)^2+(14-15.4)^2`

`\sum_(i=1)^(n)(x-\overline{x})^(2)=82.4`

Substitute back,

`s=\sqrt{(1)/(9)(82.4)}`

`s=√(9.155)`

`s=3.025`

The Standard deviation of random sample 3.025 is less than 5.8 the standard deviation of population.