Question

​Recently, a random sample of 13-18 year olds was​ asked, "How much do you currently have in​ savings?" The data in the table represent the responses to the survey. Approximate the mean and standard deviation amount of savings.

Answer 1

Answer:

μ = 235.38

σ = 234.54

Step-by-step explanation:

Assuming the table is as follows:

$\left[\begin{array}{cc}Savings&Frequency\\\ 0-\ 199&339\\\ 200-\ 399&86\\\ 400-\ 599&55\\\ 600-\ 799&18\\\ 800-\ 999&11\\\ 1000-\ 1199&8\\\ 1200-\ 1399&3\end{array}\right]$

This is an example of grouped data, where a range of values is given rather than a single data point.  First, find the total frequency.

n = 339 + 86 + 55 + 18 + 11 + 8 + 3

n = 520

The mean is the expected value using the midpoints of each range.

μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520

μ = 122400 / 520

μ = 235.38

The variance is:

σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)

σ² = 55009.7

The standard deviation is:

σ = 234.54

Answer 2

The mean of a distribution is its average, while the standard deviation is the summary of how far each data in the dataset, is to the mean. The mean and the standard deviation are 234.88 and 234.31 respectively.

First, we calculate the class midpoint (x).

This is the average of the class interval.

From the attached table, we have:

$x_1 = \frac{0 + 199}{2} = \frac{199}{2} = 99.5$

$x_2 = \frac{200 + 399}{2} = \frac{599}{2} = 299.5$

...

$x_7 = \frac{1200 + 1399}{2} = \frac{2599}{2} = 1299.5$

So, the new table becomes:

$\begin{array}{cc}Savings (x) &Frequency (f)\\99.5&339&299.5&86&499.5&55&699.5&18&899.5&11&1099.5&8&1299.5&3\end{array}$

The mean of the distribution is calculated as follows:

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{339 \times 99.5 + 86 \times 299.5 + 55 \times 499.5 + 18 \times 699.5 +11 \times 899.5 + 8 \times 1099.5 + 3 \times 1299.5}{339 + 86 + 55 + 18 + 11 + 8 + 3}$

$\bar x = \frac{122140}{520}$

$\bar x = 234.88$

The standard deviation is calculated as follows:

$\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}$

This becomes

$\sigma = \sqrt{\frac{339 \times (99.5 - 234.88)^2+ 86 \times (299.5 - 234.88)^2+...........+ 3 \times (1299.5- 234.88)^2}{339 + 86 + 55 + 18 + 11 + 8 + 3}}$

$\sigma = \sqrt{\frac{28548923.088}{520}}$

$\sigma = \sqrt{54901.7751692}$

$\sigma = 234.31$

Hence, the mean and the standard deviation are 234.88 and 234.31 respectively.

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