QUESTION IMAGE
Question
in a city with three high schools, all the ninth - graders took a standardized test, with these results:
| high school | mean score on test | number of ninth - graders |
|---|---|---|
| central city | 91 | 321 |
| lincoln high | 67 | 173 |
the citys pr manager, who never took statistics, claimed the mean score of all ninth - graders in the city was 79.3. of course, that is incorrect. what is the mean score for all ninth - graders in the city? round to one decimal place.
mean of all ninth - graders scores =
Step1: Recall the weighted - mean formula
The formula for the weighted mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}f_{i}}{\sum_{i = 1}^{n}f_{i}}$, where $x_{i}$ is the mean of the $i$-th group and $f_{i}$ is the frequency (number of observations) of the $i$-th group.
Step2: Identify the values of $x_{i}$ and $f_{i}$
For Lincoln High: $x_1 = 67$, $f_1=173$; for Central City: $x_2 = 91$, $f_2 = 321$; for Glenwood: $x_3 = 80$, $f_3=255$.
Step3: Calculate the numerator $\sum_{i = 1}^{3}x_{i}f_{i}$
$x_1f_1+x_2f_2+x_3f_3=67\times173 + 91\times321+80\times255$
$=11591+29211 + 20400$
$=61202$.
Step4: Calculate the denominator $\sum_{i = 1}^{3}f_{i}$
$f_1 + f_2+f_3=173 + 321+255=749$.
Step5: Calculate the weighted - mean
$\bar{x}=\frac{61202}{749}\approx81.7$.
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$81.7$