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Data Distributions and Statistics: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A squared-distance score

The total squared distance of a data set from a proposed center c is $S(c)=4(c-7)^2+20$ . Find the value of c that minimizes the score.
1. Hint 1
  The squared term is zero or positive for every real c.
2. Hint 2
  Choose c so that the squared term vanishes.
Answer

$c=7$ .

Full solution

The smallest possible value of $(c-7)^2$ is zero, achieved exactly when $c=7$ .
Therefore
$S(7)=20.$
Every other c adds a positive amount to $20$ , so $7$ is the unique minimizing center.

Answer

$c=7$ .

Key idea

The vertex of a total squared-distance quadratic identifies the minimizing center.
Problem 1: Mark complete
Problem 2 The data display

Read the five-number summary from the box plot in the figure. The separate point belongs to the data set.

The box plot.

Text description of this figure

A horizontal box plot drawn above a number axis labeled Value, running from 0 to 20 with a tick at every whole number and a label at every even number. The box stretches from 5 to 9, with a vertical median line inside it at 7. The left whisker runs from the box down to 2 and the right whisker from the box up to 11, each ending in a short vertical cap. Well to the right, a single filled point sits on its own at 18. No value is written on the plot.
1. Hint 1
  The five-number summary includes the actual minimum and maximum, even when an extreme is plotted separately.
2. Hint 2
  Read the two box edges, its middle line, and the smallest and largest plotted values.
Answer

Minimum 2, first quartile 5, median 7, third quartile 9, maximum 18.

Full solution

The left whisker ends at $2$ , the box starts at $5$ , its median line is at $7$ , and the box ends at $9$ .
The separate point at $18$ is the largest value.
The right whisker ends at $11$ , but it is not the maximum because the value $18$ lies beyond it.

Answer

Minimum 2, first quartile 5, median 7, third quartile 9, maximum 18.

Key idea

A five-number summary uses the largest observation, including an outlier beyond a box plot's whisker.
Problem 2: Mark complete
Problem 3 A deviation record

A sample of four observations has deviations $-2,-1,0,3$ from its own mean. Find the sample variance.
1. Hint 1
  Sample variance averages squared deviations with divisor n minus 1.
2. Hint 2
  Square the four recorded deviations and divide their total by three.
Answer

$s^2=\frac{14}{3}$ .

Full solution

The squared deviations total
$4+1+0+9=14.$
There are four observations, so the sample divisor is $4-1=3$ .
Thus
$s^2=\frac{14}{3}.$
The deviations sum to zero, consistent with being measured from their mean.

Answer

$s^2=\frac{14}{3}$ .

Key idea

Sample variance divides the sum of squared deviations by one less than the sample count.
Problem 3: Mark complete
Problem 4 Four observations

The data are $1,1,4,6$ . Find the value c minimizing total squared distance and the minimum total. Show algebraically that no other c gives a lower total.
1. Hint 1
  Expand the four squared distances as a quadratic in c.
2. Hint 2
  Complete its square; the remaining constant is the minimum.
Answer

$c=3$ ; minimum total squared distance 18.

Full solution

Expanding the four terms gives
$S(c)=4c^2-24c+54.$
Completing the square yields
$S(c)=4(c-3)^2+18.$
The squared term is zero or positive, so the minimum is $18$ , attained only at $c=3$ .
The mean is $(1+1+4+6)/4=3$ , and the squared deviations $4,4,1,9$ sum to $18$ .

Answer

$c=3$ ; minimum total squared distance 18.

Key idea

Completing the square proves that the mean uniquely minimizes total squared distance.
Problem 4: Mark complete
Problem 5 A collection point

Five pickup locations lie at positions 2, 6, 8, 12, and 14 km along a straight road. A collection point can be placed anywhere on the road. Find the position minimizing the sum of its distances to the five locations and the minimum total distance. Justify the choice.
1. Hint 1
  Pair the outermost locations, then the next outermost locations.
2. Hint 2
  A point between a pair contributes their fixed gap; the unpaired middle location determines the best position.
Answer

Position 8 km; minimum total distance 18 km.

Full solution

For the outer pair at $2$ and $14$ , the total distance is at least $12$ km.
For the pair at $6$ and $12$ , it is at least $6$ km.
Position $8$ lies inside both pairs and also makes the distance to the middle pickup zero.
The total is therefore
$12+6+0=18.$
Moving away from $8$ adds a positive distance to the middle pickup and cannot lower either pair below its gap, so $8$ is the unique minimizer.

Answer

Position 8 km; minimum total distance 18 km.

Key idea

The median minimizes total absolute distance; for an odd count it is the unique minimizer, and for an even count every value between the two middle observations minimizes it.
Problem 5: Mark complete
Problem 6 A distribution to summarize

The histogram in the figure shows daily rainfall totals in millimeters. Recommend measures of center and spread for a typical day. Identify the bin containing the median, and explain whether the exact median can be recovered.

Daily rainfall totals.

Text description of this figure

A histogram with the horizontal axis labeled Daily rainfall (mm), marked every 10 millimeters from 0 to 60, and the vertical axis labeled Frequency, marked every 2 from 0 to 12. Six adjoining bars, each 10 millimeters wide, have heights 3 for 0 to 10, 12 for 10 to 20, 8 for 20 to 30, 3 for 30 to 40, 1 for 40 to 50, and 1 for 50 to 60. The tallest bar is near the left, and the bars trail off in a long, low stretch to the right. No center, spread or shape is labeled.
1. Hint 1
  Look for a long tail that would pull statistics based on squared distance.
2. Hint 2
  Pair a resistant center with a resistant spread when a long tail is present.
Answer

Median and IQR; the median lies in the 10-to-20 mm bin, but its exact value cannot be recovered from the histogram.

Full solution

Most days lie in the lower rainfall bins, and a thin tail extends toward heavy rainfall.
This is a right-skewed display, and the mean and standard deviation respond strongly to those large values, so the median and IQR describe a typical day more reliably.

The bars hold $3,12,8,3,1,1$ days, $28$ in all, with cumulative counts $3,15,23,26,27,28$ .
The median is the average of the $14$ th and $15$ th values, and both lie in the 10-to-20 mm bin.
A histogram records only how many values fall in each bin, not where they lie inside it, so the exact median cannot be recovered.

Answer

Median and IQR; the median lies in the 10-to-20 mm bin, but its exact value cannot be recovered from the histogram.

Key idea

This histogram supports the median and IQR, and here it locates the median's bin without fixing its exact value.
Problem 6: Mark complete
Problem 7 An unknown upper value

The sorted data are $2,4,5,6,8,9,10,12,t$ , where $t\ge12$ . Exclude the median from both halves when finding quartiles. For which values of t is t an outlier under the $1.5$ times IQR rule?
1. Hint 1
  The middle observation is excluded, leaving four observations in each half.
2. Hint 2
  Check whether either quartile depends on t before forming the upper fence.
Answer

$t>20.75$ .

Full solution

The median is $8$ .
The lower half has middle values $4,5$ , so $Q_1=4.5$ .
The upper half has middle values $10,12$ , so $Q_3=11$ , independent of t.
Hence
$\begin{aligned}\mathrm{IQR}&=11-4.5 \\ \mathrm{IQR}&=6.5.\end{aligned}$
The upper fence is
$11+1.5(6.5)=20.75.$
Values strictly beyond it are flagged, so $t>20.75$ .
Equality is not an outlier.

Answer

$t>20.75$ .

Key idea

When an extreme stays outside the quartile calculation, its outlier threshold can remain fixed as it moves.
Problem 7: Mark complete
Problem 8 A spread report

A population has deviations $-4,-1,2,3$ from its mean. A student averages them and reports zero spread. Is this a valid conclusion? Find the population variance to support your answer. Explain why deviations from the mean sum to zero for every nonempty data set.
1. Hint 1
  Signed deviations cancel even when the values are different.
2. Hint 2
  Square the deviations to remove the cancellation, then use the population divisor.
Answer

No; the population variance is $7.5$ . Deviations sum to zero because $n$ copies of the mean total the same as the data.

Full solution

The signed deviations sum to zero, but that does not mean they individually vanish.
Their squares total
$16+1+4+9=30.$
The four observations form the population, so
$\begin{aligned}\sigma^2&=\frac{30}{4} \\ \sigma^2&=7.5.\end{aligned}$
The positive variance confirms nonzero spread; the signed average hid it through cancellation.
In general, subtracting the mean once per observation subtracts the same total as the sum of the observations, which explains why their signed deviations sum to zero.

Answer

No; the population variance is $7.5$ . Deviations sum to zero because $n$ copies of the mean total the same as the data.

Key idea

Averaging signed deviations hides spread because deviations from the mean sum to zero.
Problem 8: Mark complete
Problem 9 One sample's variance

Six independent random observations come from one population with variance $12$ . Their squared deviations from their sample mean total $85$ . A student calculates the sample variance and says that any answer above $12$ would prove the correction failed. Find the sample variance and assess the claim. Explain what the correction accomplishes across repeated samples.
1. Hint 1
  Distinguish one sample's estimate from the average of the estimates over many samples.
2. Hint 2
  Use the sample divisor, one less than the number of observations.
Answer

$s^2=17$ ; the claim is false, since the correction makes the estimates right on average across repeated samples, and a single estimate can exceed $12$ .

Full solution

The sample variance divides the squared-deviation total by $n-1=5$ :
$s^2=\frac{85}5$
which is $17$ .

Measuring from the sample mean makes the squared-deviation total no larger than measuring from the population mean, with equality when the two means agree, so dividing by $n$ would underestimate the variance on average.
Dividing by $n-1$ removes that bias: over many independent random samples the estimates average to the population variance.
Any single estimate still varies, so $17>12$ shows ordinary sampling variation, not a failure of the correction.

Answer

$s^2=17$ ; the claim is false, since the correction makes the estimates right on average across repeated samples, and a single estimate can exceed $12$ .

Key idea

For independent random samples, the n minus 1 divisor makes the sample variance correct on average, not in every sample.
Problem 9: Mark complete
Problem 10 An observation to investigate

A measurement lies beyond an IQR fence. A student concludes it must be a recording mistake and should be deleted. Is that conclusion justified by the fence alone? Explain what the outlier flag establishes.
1. Hint 1
  A fence is a convention for flagging unusually distant observations.
2. Hint 2
  Consider whether an unusual value might still be an accurate measurement.
Answer

No; the observation warrants investigation, but the fence alone does not establish an error.

Full solution

The upper and lower fences are defined using a chosen multiplier:
$Q_3+1.5\,\mathrm{IQR}$
and
$Q_1-1.5\,\mathrm{IQR}.$
Falling beyond one shows that the observation is far from the middle half under this convention.
It supplies no evidence by itself that the instrument or record is wrong.
The value could be valid, so its source should be investigated before deciding whether a correction is warranted.

Answer

No; the observation warrants investigation, but the fence alone does not establish an error.

Key idea

An IQR outlier flag identifies an observation to investigate rather than proving it should be removed.
Problem 10: Mark complete

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