The amount of money in an average teenager’s savings account in the U.S. is normally distributed with a mean of $967 and a standard deviation of $122. Use a spreadsheet program to find the percentage of teens with more than $1,200 in their savings account.
To find the percentage of teenagers with more than $1,200 in their savings account, we need to calculate the z-score for $1,200 and then find the percentage of data points above that z-score in a standard normal distribution.
The formula to calculate the z-score is:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- X = $1,200
- μ (mean) = $967
- σ (standard deviation) = $122
Substitute the values into the formula:
\[ z = \frac{1,200 - 967}{122} \]
\[ z = \frac{233}{122} \]
\[ z ≈ 1.91 \]
Next, we can use a standard normal distribution table or a spreadsheet program to find the percentage of data points above a z-score of 1.91.
Using a spreadsheet program, we can use the function NORM.S.DIST(z, TRUE) or NORM.S.DIST(z, 1) to calculate this.
\[ NORM.S.DIST(1.91, TRUE) \]
This will give us the percentage of teenagers with more than $1,200 in their savings account.
The result is approximately 3.19%. Therefore, around 3.19% of teenagers have more than $1,200 in their savings account in the U.S.