The marks scored by some students are summarized in the following table:
| Mark, $x$ | Frequency |
|---|
| $0 < x \leqslant 5$ | $5$ |
| $5 < x \leqslant 10$ | $7$ |
| $10 < x \leqslant 15$ | $8$ |
| $15 < x \leqslant 20$ | $5$ |
- Find the modal class.
- Find an estimate for the mean.
- Explain why this is only an estimate.
- $10 < x \leqslant15$
- $10.1$
- The distribution of the marks within each class is unknown.
A group of students took a test. The mean mark for $12$ boys was $18$, and the mean mark for $8$ girls was $16.5$. What is the mean mark for all $20$ students?
$17.4$
The salary of six teachers are (in thousands of pounds): $$24 \quad 29 \quad 33 \quad 35 \quad 62$$ - Find the mean and median salary of the teachers.
- The school wants to give the teachers an idea of the average amount they earn. Which of these two averages is more useful?
- Mean is $36.6$, median is $33$
- The median is more useful because the mean is affected by the extreme data point (62)
The numbers $a$, $b$ and $c$ have a median of $4$, a range of $7$, and a mean of $5.2$. Find the median, range and mean of $2a+1$, $2b+1$ and $2c+1$.
Median: $9$, Range: $14$, Mean: $11.4$
Given $$\sum x = 50 \quad \sum x^2 = 410 \quad n = 10$$ find the mean and the standard deviation.
Mean: $5$ Standard Deviation: $4$
The manager of a clothes shop recorded the size of each dress sold one morning to help decide what size dresses to order in the future: $$2\ 4\ 4\ 8\ 8\ 8\ 10\ 10\ 12\ 12\ 16\ 18$$ - Find the mean dress size sold and the standard deviation.
- Dress sizes are only in even numbers. Explain why the mean dress size is not very useful, and suggest a more useful average for the manager to use.
- Mean is $9.3$, standard deviation is $4.57$
- The manager cannot order dresses in this size. A more useful average is the mode.
80 students were asked to choose a number, $x$, between $0$ and $100$. | $x$ | Frequency |
|---|
| $0 < x \leqslant 25$ | $23$ |
| $25 < x \leqslant 50$ | $18$ |
| $50 < x \leqslant 75$ | $14$ |
| $75 < x \leqslant 100$ | $25$ |
Use interpolation to find an estimate for the interquartile range of the numbers chosen. $80 - 21.7 = 58.3$
The amount of time 200 students take to get to school in the mornings, $m$ minutes, were collected:
| $m$ | Frequency |
|---|
| $0 < m \leqslant 10$ | $34$ |
| $10 < m \leqslant 20$ | $53$ |
| $20 < m \leqslant 30$ | $76$ |
| $30 < m \leqslant 40$ | $37$ |
Calculate the $10$% to $90$% interpercentile range. $34.6 - 5.9 = 28.7$
The heights of 10 girls, in cm, have the following summary statistics: $$\sum x = 1570 \quad \sum x^2 = 246900$$ The heights of 8 boys have the following summary statistics: $$\sum x = 1335 \quad \sum x^2 = 223275$$ Find the mean and standard deviation of all 18 children.
Mean: $161$, Standard Deviation: $8.63$
The number of detentions, $d$, 100 students received last term were recorded:
Calculate the mean and standard deviation. Mean: $0.43$, standard deviation: $0.828$
The amount of pocket money, $p$ pounds, 80 students receive per week was recorded:
| $p$ | Frequency |
|---|
| $0 < p \leqslant 5$ | $28$ |
| $5 < p \leqslant 10$ | $24$ |
| $10 < p \leqslant 15$ | $17$ |
| $15 < p \leqslant 20$ | $11$ |
Estimate the number of students who earn more than 1 standard deviation above the mean amount. Mean: $8.19$, standard deviation: $5.23$ <br> $16$ students.
The heights of a group of students, $x$ cm, were coded using the formula $$y = 5(x-165)$$ The summary statistics for $y$ are as follows: $$n = 10 \quad \sum y = 125 \quad \sum y^2 = 11875$$ - Find the mean and standard deviation of $y$.
- Hence find the mean and standard deviation of $x$.
- Mean: $12.5$, standard deviation: $32.1$
- Mean: $167.5$, standard deviation: $6.42$
The lengths of time each student in a class of $25$ spent chatting in a lesson were recorded.
The lower quartile of times was found to be $5$ minutes, and the upper quartile was found to be $12$ minutes.
An outlier is defined as any data point which lies more than $1$ interquartile range above the upper quartile, or below the lower quartile.
What are the largest and smallest amounts of time a student can chat in a lesson without being an outlier?
More than $19$ minutes
The number of after school clubs, $x$, at 100 schools are recorded.
| $x$ | Frequency |
|---|
| $0 < x \leqslant 3$ | $15$ |
| $3 < x \leqslant 6$ | $48$ |
| $6 < x \leqslant 9$ | $37$ |
The data is coded using $$y = \frac{x - 3}{10}$$ Find an estimate for the mean and standard deviation of $y$. $\bar{y} = 0.216$, $\sigma_y = 0.206$
The scores of 10 people who sat a maths test, $x$, are summarized: $$\sum x = 608 \quad \sum x^2 = 40034$$ An outlier is defined as any data point which lies outside of $2$ standard deviations of the mean. - Find the smallest and highest test score that would not be considered an outlier.
- One of the 10 people who sat the test was the teacher, who scored $100$. Clean the data and find the mean and standard deviation of the 9 students who sat the test.
- Mean: $60.8$, standard deviation: $17.5$, outliers: $25.8$ to $95.8$
- Mean: $56.4$, standard deviation: $12.3$
A school experimented with allowing half a class of $20$ students to use their phones in lessons. The scores of the students who used phones, $x$, and who did not use their phones, $y$, in an end of year test are summarized: $$\sum x = 557 \quad \sum x^2 = 39125$$ $$\sum y = 602 \quad \sum y^2 = 37934$$ By finding the mean and standard deviation, compare the effectiveness of the use of phones in lessons on test results.
$\bar{x} = 55.7$, $\sigma_x = 28.5$ <br> $\bar{y} = 60.2$, $\sigma_y = 13.0$ <br> On average, the students who used their phones did worse <br> The students who used phones had a larger variation in how well they did.
The amount of time, $h$ hours, students spend on social media on a school night were recorded.
| $h$ | Frequency |
|---|
| $0 < h \leqslant 1$ | $12$ |
| $1 < h \leqslant 2$ | $31$ |
| $2 < h \leqslant 3$ | $18$ |
| $3 < h \leqslant 4$ | $14$ |
An outlier is defined as any data point which lies outside of $1$ standard deviation of the mean.
Estimate the number of outliers in this data. $27$
A student calculated some statistics for a recent test their class of $20$ students did, and found that the mean mark was $62$, and the standard deviation was $28$. One student found that a page of their test was unmarked, and their score was increased from $46$ to $64$.
Find the new mean and standard deviation of the marks.
Mean: $62.9$, standard deviation: $27.8$