Question

1. which of the following statements is (are) true?

i. the median is resistant to extreme values.
ii. the mean is resistant to extreme values.
iii. the standard deviation is resistant to extreme values.
(a) i only (d) ii and iii only
(b) ii only (e) i and iii only
(c) iii only

1. a set of data has a mean that is much larger than the median. which of the following statements is most consistent with this information?

(a) the distribution is symmetric
(b) the distribution is skewed left
(c) the distribution is skewed right
(d) the distribution is bimodal
(e) the data set probably has a few low outliers

1. the mean age of teachers at high school a is 34.2 and the mean age of teachers at high school b is 39.13. the sum of all the ages at high school a is 1197 and the sum of all the ages of high school b is 9... many teachers are in each high school? (formula for mean: \\(\bar{x} = \frac{\sum x_i}{n}\\))

(a) 35 teachers in high school a and 23 teachers in high school b
(b) 30 teachers in high school a and 19 teachers in high school b
(c) 25 teachers in high school a and 17 teachers in high school b
(d) 19 teachers in high school a and 30 teachers in high school b
(e) 23 teachers in high school a and 35 teachers in high school b

1. the variance of 40 values is 100. what is the standard deviation of that set of data?

(hints: variance: \\(\frac{\sum((\bar{x}-x_i)^2)}{n-1}\\) standard deviation: \\(\sqrt{\frac{\sum((\bar{x}-x_i)^2)}{n-1}}\\))
(a) 10 (d) 25
(b) 0.4 (e) 2.5

Explanation:

Question 5

• For statement I: The median is the middle value, so extreme values (outliers) don't affect its position much, making it resistant.
• For statement II: The mean is calculated as the sum of all values divided by the number of values. Extreme values can pull the mean up or down, so it is not resistant.
• For statement III: The standard deviation depends on the mean (since it measures the spread from the mean) and on the individual data points. Extreme values will affect the mean and the squared differences, so standard deviation is not resistant.

So only statement I is true.

• In a symmetric distribution, the mean and median are approximately equal.
• In a left - skewed (negatively skewed) distribution, the mean is less than the median because the low outliers pull the mean down.
• In a right - skewed (positively skewed) distribution, the mean is greater than the median because the high outliers pull the mean up.
• A bimodal distribution has two peaks but doesn't directly relate to the relationship between mean and median.
• Low outliers would make the mean less than the median (left - skewed), not greater.

Since the mean is much larger than the median, the distribution is skewed right.

Step 1: Find the number of teachers in High School A

We know that the formula for the mean $\bar{x}=\frac{\sum x_{i}}{n}$, where $\bar{x}$ is the mean, $\sum x_{i}$ is the sum of the values, and $n$ is the number of values. For High School A, $\bar{x} = 34.2$ and $\sum x_{i}=1197$. We can solve for $n$ (number of teachers) by rearranging the formula: $n=\frac{\sum x_{i}}{\bar{x}}$.
Substitute the values: $n=\frac{1197}{34.2}=35$? Wait, no, $34.2\times35 = 1197$? Wait, $34.2\times35=34.2\times(30 + 5)=34.2\times30+34.2\times5 = 1026+171 = 1197$. Wait, but let's check the options. Wait, maybe there is a typo in the sum for High School B. Assuming the sum for High School B is $9\times39.13$? Wait, no, let's check the options. Let's calculate for High School A: $n=\frac{1197}{34.2}=35$? Wait, no, $34.2\times35 = 1197$, but let's check the options. Option A: 35 in A and 23 in B. Let's check High School B: If $n = 23$ and $\bar{x}=39.13$, then $\sum x_{i}=\bar{x}\times n=39.13\times23 = 39.13\times(20 + 3)=782.6+117.39 = 899.99\approx900$. So that makes sense. Wait, maybe the sum for High School B is 900 (a typo in the image). So for High School A: $n=\frac{1197}{34.2}=35$? Wait, no, $34.2\times35 = 1197$, yes. For High School B: $n=\frac{900}{39.13}\approx23$. So the number of teachers in High School A is 35 and in High School B is 23.

Step 2: Verify the calculation for High School A

Using the mean formula $\bar{x}=\frac{\sum x_{i}}{n}$, we have $n=\frac{\sum x_{i}}{\bar{x}}=\frac{1197}{34.2}=35$.

Step 3: Verify the calculation for High School B

Let the sum of ages in High School B be $S$. Then $n=\frac{S}{\bar{x}}$. If $n = 23$ and $\bar{x}=39.13$, then $S=39.13\times23\approx900$ (which matches the likely intended sum, considering the typo).

Answer:

A. I only

Question 6