Answer:
Explained below.
Step-by-step explanation:
(10)
The data set is:
S = {124, 94, 129, 109, 114}
The mean and standard deviation are:
![\bar x=\frac{1}{n}\sum x=\frac{1}{5}\times [124+94+...+114]=114\\\\s=\sqrt{\frac{1}{n-1}\sum ( x-\bar x)^{2}}](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%20x%3D%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5B124%2B94%2B...%2B114%5D%3D114%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28%20x-%5Cbar%20x%29%5E%7B2%7D%7D)
![=\sqrt{\frac{1}{5-1}\times [(124-114)^{2}+(94-114)^{2}+...+(114-114)^{2}]}\\=\sqrt{\frac{750}{4}}\\=13.6931\\\approx 13.69](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B5-1%7D%5Ctimes%20%5B%28124-114%29%5E%7B2%7D%2B%2894-114%29%5E%7B2%7D%2B...%2B%28114-114%29%5E%7B2%7D%5D%7D%5C%5C%3D%5Csqrt%7B%5Cfrac%7B750%7D%7B4%7D%7D%5C%5C%3D13.6931%5C%5C%5Capprox%2013.69)
The correct option is B.
(11)
According to the Empirical 95% of the data for a Normal distribution are within 2 standard deviations of the mean.
So, the adult male's height is in the same range as about 95% of the other adult males whose heights were measured.
The correct option is B.
(12)
Let the score be <em>X</em>.
Given:
μ = 100
σ = 26
![X=\mu-2\sigma](https://tex.z-dn.net/?f=X%3D%5Cmu-2%5Csigma)
![=100-(2\times 26)\\=100-52\\=48](https://tex.z-dn.net/?f=%3D100-%282%5Ctimes%2026%29%5C%5C%3D100-52%5C%5C%3D48)
The correct option is B.
(13)
Let <em>X</em> be the prices of a certain model of new homes.
Given: ![X\sim N(150000, 2300^{2})](https://tex.z-dn.net/?f=X%5Csim%20%20N%28150000%2C%202300%5E%7B2%7D%29)
Compute the percentage of buyers who paid between $147,700 and $152,300 as follows:
![P(147700](https://tex.z-dn.net/?f=P%28147700%3CX%3C152300%29%3DP%28%5Cfrac%7B147700-150000%7D%7B2300%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3C%5Cfrac%7B152300-150000%7D%7B2300%7D%29)
According to the 68-95-99.7, 68% of the data for a Normal distribution are within 1 standard deviations of the mean.
The correct option is D.
(14)
Compute the percentage of buyers who paid more than $154,800 as follows:
![P(X>154800)=P(\frac{X-\mu}{\sigma}>\frac{154800-150000}{2400})](https://tex.z-dn.net/?f=P%28X%3E154800%29%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B154800-150000%7D%7B2400%7D%29)
![=P(Z>2)\\=0.975\\](https://tex.z-dn.net/?f=%3DP%28Z%3E2%29%5C%5C%3D0.975%5C%5C)
According to the 68-95-99.7, 95% of the data for a Normal distribution are within 2 standard deviations of the mean. Then the percentage of data above 2 standard deviations of the mean will be 97.5% and below 2 standard deviations of the mean will be 2.5%.
The correct option is D.
(15)
The <em>z</em>-score is given as follows:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)