Question

1. another college is asked and says students pay $15,000. what is the effect on the mean and standard deviation?
2. calculate the z - score for the $15,000 value.

Explanation:

1. Effect on mean and standard - deviation (assuming we have an existing data - set of college tuition values):

• Effect on the mean:
• # Explanation:

Step1: Recall the formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. If we add a new data - point $x_{new}=15000$ to a data - set of size $n$ with a current sum of values $\sum_{i = 1}^{n}x_{i}$, the new sum is $\sum_{i = 1}^{n}x_{i}+x_{new}$ and the new number of data - points is $n + 1$. The new mean $\bar{x}_{new}=\frac{\sum_{i = 1}^{n}x_{i}+x_{new}}{n + 1}$. If $15000$ is greater than the original mean, the new mean will increase; if it is less than the original mean, the new mean will decrease.

Let the original sum of $n$ data - points be $S=\sum_{i = 1}^{n}x_{i}$ and the original mean be $\bar{x}=\frac{S}{n}$. The new mean $\bar{x}_{new}=\frac{S + 15000}{n+1}$.
For example, if the original $n = 5$ data - points have a sum $S = 50000$ (so the original mean $\bar{x}=\frac{50000}{5}=10000$), then the new sum is $S+15000=50000 + 15000=65000$ and the new number of data - points is $n + 1=6$. The new mean is $\bar{x}_{new}=\frac{65000}{6}\approx10833.33$, which is an increase.

Step2: Recall the formula for the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. Adding a new data - point changes both the mean and the sum of squared deviations. If the new data - point $x_{new}=15000$ is far from the original mean, it will increase the standard deviation. If it is close to the original mean, it will have a small effect on the standard deviation.

The new standard deviation calculation will involve the new mean $\bar{x}_{new}$ and the new sum of squared - deviations $\sum_{i = 1}^{n}(x_{i}-\bar{x}_{new})^{2}+(x_{new}-\bar{x}_{new})^{2}$.

• Answer: The effect on the mean depends on whether $15000$ is greater or less than the original mean (increases if greater, decreases if less). The effect on the standard deviation depends on how far $15000$ is from the original mean (increases if far, has little effect if close).

1. Calculating the z - score:

• # Explanation:

Step1: Recall the formula for the z - score $z=\frac{x-\mu}{\sigma}$, where $x = 15000$, $\mu$ is the population mean, and $\sigma$ is the population standard deviation. If we are using sample statistics, we can use $\bar{x}$ for the mean and $s$ for the standard deviation.

Let's assume the population mean $\mu$ and standard deviation $\sigma$ are known. For example, if $\mu = 12000$ and $\sigma = 2000$.

Step2: Substitute the values into the formula.

$z=\frac{15000 - 12000}{2000}=\frac{3000}{2000}=1.5$

• # Answer: $z=\frac{15000-\mu}{\sigma}$ (in general), and if $\mu = 12000$ and $\sigma = 2000$, then $z = 1.5$

Answer:

1. Effect on mean and standard - deviation (assuming we have an existing data - set of college tuition values):

• Effect on the mean:
• # Explanation:

Step1: Recall the formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. If we add a new data - point $x_{new}=15000$ to a data - set of size $n$ with a current sum of values $\sum_{i = 1}^{n}x_{i}$, the new sum is $\sum_{i = 1}^{n}x_{i}+x_{new}$ and the new number of data - points is $n + 1$. The new mean $\bar{x}_{new}=\frac{\sum_{i = 1}^{n}x_{i}+x_{new}}{n + 1}$. If $15000$ is greater than the original mean, the new mean will increase; if it is less than the original mean, the new mean will decrease.

Let the original sum of $n$ data - points be $S=\sum_{i = 1}^{n}x_{i}$ and the original mean be $\bar{x}=\frac{S}{n}$. The new mean $\bar{x}_{new}=\frac{S + 15000}{n+1}$.
For example, if the original $n = 5$ data - points have a sum $S = 50000$ (so the original mean $\bar{x}=\frac{50000}{5}=10000$), then the new sum is $S+15000=50000 + 15000=65000$ and the new number of data - points is $n + 1=6$. The new mean is $\bar{x}_{new}=\frac{65000}{6}\approx10833.33$, which is an increase.

Step2: Recall the formula for the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. Adding a new data - point changes both the mean and the sum of squared deviations. If the new data - point $x_{new}=15000$ is far from the original mean, it will increase the standard deviation. If it is close to the original mean, it will have a small effect on the standard deviation.

The new standard deviation calculation will involve the new mean $\bar{x}_{new}$ and the new sum of squared - deviations $\sum_{i = 1}^{n}(x_{i}-\bar{x}_{new})^{2}+(x_{new}-\bar{x}_{new})^{2}$.

• Answer: The effect on the mean depends on whether $15000$ is greater or less than the original mean (increases if greater, decreases if less). The effect on the standard deviation depends on how far $15000$ is from the original mean (increases if far, has little effect if close).

1. Calculating the z - score:

• # Explanation:

Step1: Recall the formula for the z - score $z=\frac{x-\mu}{\sigma}$, where $x = 15000$, $\mu$ is the population mean, and $\sigma$ is the population standard deviation. If we are using sample statistics, we can use $\bar{x}$ for the mean and $s$ for the standard deviation.

Let's assume the population mean $\mu$ and standard deviation $\sigma$ are known. For example, if $\mu = 12000$ and $\sigma = 2000$.

Step2: Substitute the values into the formula.

$z=\frac{15000 - 12000}{2000}=\frac{3000}{2000}=1.5$

• # Answer: $z=\frac{15000-\mu}{\sigma}$ (in general), and if $\mu = 12000$ and $\sigma = 2000$, then $z = 1.5$