the following data represents the percentages of family income allocated to groceries for a sample of 50 shoppers:

percentage of family income #of shoppers
10-19 6
20-29 14
30-39 16
40-49 11
50-59 3
1.calculate the value of mean
1.1 meadian percentage of family income spent on groceries
1.12 determine the standard deviation and comment on the spread

If you are expected to do this by hand, start by calculating the mean. (The median is the "middle" of the distribution.) For grouped data, the mean can be calculated using the following formula:
mean = (Efx)/n
...where n represents the total frequencies (50 in this case), x represents the midpoint of each class interval (which is found by adding the lowest and the highest number in the interval together and dividing by 2), and E represents the sum of each midpoint times each frequency f. It might be helpful to set up a table of midpoints for each interval, then multiply each midpoint by its respective frequency (number in each interval), then add those numbers to get a total. Then divide the total by n.

Standard deviation is the square root of the variance. For grouped data, the variance can be calculated using the following formula:
variance = Ef(x - mean)^2/(n-1)
...where E represents sum of frequency f times (x-mean)^2, and n = 50. (x represents the midpoint of each class interval.)
Again it might be helpful to set up tables with the following headings:

x - mean ...... (x - mean)^2 ..... f(x - mean)^2

Add up the last column for a total, then divide by 49 (which is n-1 or 50-1 = 49). When you are finished calculating the variance, take the square root of the variance and you will have your standard deviation for this group of data.
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Hope this will help get you started on these problems,

To calculate the mean, we need to find the sum of the products of each midpoint and its respective frequency.

Step 1: Create a table showing the midpoints, frequencies, and products.

Midpoint (x) Frequency (f) fx
10-19 6 15
20-29 14 42
30-39 16 54
40-49 11 44
50-59 3 18

Step 2: Calculate the sum of the products (Σfx).

Σfx = 15 + 42 + 54 + 44 + 18 = 173

Step 3: Calculate the mean.

Mean = Σfx/n
= 173/50
≈ 3.46

So, the mean percentage of family income allocated to groceries is approximately 3.46%.

To calculate the median, we need to find the middle value of the distribution. Since we have grouped data, we'll use the cumulative frequency to determine the median class interval.

Step 1: Calculate the cumulative frequency by adding up the frequencies.

Cumulative Frequency:
10-19: 6
20-29: 6 + 14 = 20
30-39: 20 + 16 = 36
40-49: 36 + 11 = 47
50-59: 47 + 3 = 50

Step 2: Find the median class interval.
The median class interval is where the cumulative frequency becomes greater than or equal to n/2 (25 in this case).

The median class interval is 30-39.

Step 3: Calculate the median.
To calculate the median, we need to use the median formula:

Median = Lower Bound of Median Class + ((n/2 - Cumulative Frequency of the Class) / Frequency of the Median Class) * Class Width

In this case:
Lower Bound of Median Class = 30
Frequency of the Median Class = 16
Cumulative Frequency of the Class = 20

Median = 30 + ((25 - 20) / 16) * 10
= 30 + (5/16) * 10
≈ 30 + 1.563
≈ 31.563

So, the median percentage of family income spent on groceries is approximately 31.563%.

To calculate the standard deviation, we need to find the variance first.

Step 1: Create a table showing the deviations from the mean, squared deviations, and products of frequencies and squared deviations.

Deviation (x - mean) Squared Deviation (x - mean)^2 f(x - mean)^2
10-19 (10-3.46)^2 42.49316
20-29 (20-3.46)^2 310.53476
30-39 (30-3.46)^2 729.15076
40-49 (40-3.46)^2 1467.11616
50-59 (50-3.46)^2 2436.91376

Step 2: Calculate the sum of f(x - mean)^2.

Σf(x - mean)^2 = 42.49316 + 310.53476 + 729.15076 + 1467.11616 + 2436.91376
≈ 4986.2084

Step 3: Calculate the variance.

Variance = Σf(x - mean)^2 / (n - 1)
= 4986.2084 / (50 - 1)
= 4986.2084 / 49
≈ 101.55335

Step 4: Calculate the standard deviation.

Standard Deviation = √Variance
= √101.55335
≈ 10.07724

The standard deviation for this group of data is approximately 10.07724.

The standard deviation measures the spread or variability of the data. In this case, the value of approximately 10.07724 indicates that the data points are dispersed or spread out around the mean by an average of about 10.07724 percentage points. A greater standard deviation suggests a wider spread of data, while a smaller standard deviation suggests a narrower spread.