Question

12. Provide an appropriate response. Samples of size n- 240 are randomly selected from the population of numbers (0 through 20) produced by a random-number generator, and the variance is found for each sample. What is the distribution of the sample variances? O normal (approximately) O skewed to the right O skewed to the left O not enough information provided 13. Choose the correct response. ( point) Why is sampling without replacement acceptable with a large population? When a small sample is taken from a large population, the samples maintain their independence O When a large sample is take from a large population, the sample retains the characteristics of the original distribution When a sample without replacement is taken, there is no requirement to maintain the same sample size When a sample without replacement is taken, the samples will have the appropriate amount of data

Answer 1

#5 By using standard normal table (shown below), the probability that z lies between 0 and 3.01 is:

$P(0<z<3.01)=P(z<3.01)-P(z<0)=0.9987-0.5000={\color{Red} 0.4987}$

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#6 Look into standard normal table (shown below) for area closest to 0.4500, which is 0.4483. The z score corresponding to this is -0.13. Therefore, we can say, z score corresponding to P₄₅ is $z=-0.13$.

$z=\frac{x-\mu}{\sigma}\Rightarrow x=\mu&plus;z\sigma =1050-0.13(218)=1021.66\approx {\color{Red} 1021.7}$

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#8 By using standard normal table (shown below), the probability that September energy consumption level is between 1100 kWh and 1225 kWh is:

$\\P(1100<x<1225)=P\left ( \frac{1100-1050}{218}<\frac{x-\mu}{\sigma}<\frac{1225-1050}{218} \right )\\\\=P(0.23<z<0.80)=P(z<0.80)-P(z<0.23) \\\\=0.7881-0.5910={\color{Red} 0.1971}$

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#9 By using standard normal table (shown below), the probability that IQ score is between 85 and 125 is:

$\\P(85<x<125)=P\left ( \frac{85-100}{15}<\frac{x-\mu}{\sigma}<\frac{125-100}{15} \right )\\\\=P(-1.00<z<1.67)=P(z<1.67)-P(z<-1.00) \\\\=0.9525-0.1587={\color{Red} 0.7938}$

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#12 Since variance cannot be negative, the sample variance values must be 0 or above. Therefore, the distribution of sample variances must be ${\color{Red} \textup{skewed to the right}}$ .

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#13 When small sample (less than 5% of the population) is taken from the population, the samples maintain their independence. Therefore, correct answer is ${\color{Red} \textup{first option}}$ .