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Question
- the weights of male vizslas (a dog breed) aged two years and older are approximately normally distributed. a vizsla weighing 23.5 kilograms is at the 23rd percentile of vizslas. which of the following could describe the mean and standard deviation of the weights of the dogs? a) mean: 29 kg standard deviation: 2.75 kg b) mean: 27 kg standard deviation: 4.75 kg c) mean: 23.5 kg standard deviation: 4.75 kg d) mean: 22 kg standard deviation: 2.75 kg e) mean: 20 kg standard deviation: 4.75 kg 9. in 2001 in the national league of major league baseball, the league average for home runs hit per player was 11.98. the standard deviation of home runs hit per player was 11.7. which of the following is most likely true about the distribution of home runs hit per player in the national league? a) since the mean and standard deviation are nearly the same, the distribution of home runs hit is approximately normal. b) since the mean and standard deviation are nearly the same and the minimum number of home runs hit per player is zero, the distribution of home runs hit is skewed right. c) since some players hit more than 11.98 home runs and some players hit less than 11.98 home runs, the distribution of home runs hit must be approximately normal. d) since the standard deviation is smaller than the mean, the distribution of home runs hit is approximately normal. e) since the standard deviation is smaller than the mean, the distribution of home runs hit is skewed left.
Step1: Recall z - score for 23rd percentile
The z - score corresponding to the 23rd percentile can be found from the standard normal distribution table. The z - score $z$ for the 23rd percentile is approximately $z=- 0.74$.
Step2: Use the z - score formula
The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $x = 23.5$ (the value of the data - point), $\mu$ is the mean, and $\sigma$ is the standard deviation. We can rewrite it as $x=\mu+z\sigma$. Substituting $x = 23.5$ and $z=-0.74$, we get $23.5=\mu - 0.74\sigma$.
Step3: Test each option
- Option A: If $\mu = 29$ and $\sigma=2.75$, then $\mu - 0.74\sigma=29-0.74\times2.75=29 - 2.035 = 26.965
eq23.5$.
- Option B: If $\mu = 27$ and $\sigma = 4.75$, then $\mu-0.74\sigma=27-0.74\times4.75=27 - 3.515 = 23.485\approx23.5$.
- Option C: If $\mu = 23.5$ and $\sigma = 4.75$, then $\mu-0.74\sigma=23.5-0.74\times4.75=23.5 - 3.515 = 20.085
eq23.5$.
- Option D: If $\mu = 22$ and $\sigma = 2.75$, then $\mu-0.74\sigma=22-0.74\times2.75=22 - 2.035 = 19.965
eq23.5$.
- Option E: If $\mu = 20$ and $\sigma = 4.75$, then $\mu-0.74\sigma=20-0.74\times4.75=20 - 3.515 = 16.485
eq23.5$.
For the second question:
Step1: Analyze the properties of normal and skewed distributions
In a normal distribution, the mean, median, and mode are approximately equal. In a skewed - right distribution, the mean is greater than the median, and in a skewed - left distribution, the mean is less than the median. The minimum number of home - runs is 0. If the mean and standard deviation are nearly the same ($\mu = 11.98$ and $\sigma = 11.7$) and the minimum value is 0, there are likely some large values (outliers) that pull the mean up. This indicates a right - skewed distribution.
- Option A: Just because the mean and standard deviation are nearly the same does not mean the distribution is normal.
- Option B: Since the mean and standard deviation are nearly the same and the minimum number of home - runs is 0, the large values (outliers) pull the mean up, so the distribution is skewed right. This is correct.
- Option C: The fact that some players hit more and some hit less than the mean does not imply normality.
- Option D: The relationship between the mean and standard deviation alone does not determine normality.
- Option E: Since the mean is larger than the standard deviation and the minimum is 0, it is not skewed left.
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- B. Mean: 27 kg Standard Deviation: 4.75 kg
- B. Since the mean and standard deviation are nearly the same and the minimum number of home runs hit per player is zero, the distribution of home runs hit is skewed right.