# Descriptive measures

What is the percentage of people who earned between Rs 75 and Rs 125? If the given frequency table is below:

• What is "Rs", is that the same as "weekly wage"?
– Gijs
Jan 11, 2021 at 11:05
• Please read the wiki about the self-study tag and consider acting upon that: stats.stackexchange.com/tags/self-study/info Jan 11, 2021 at 11:17
• @Gijs yes 'Rs' means weekly wages Jan 11, 2021 at 11:22

Even without a self-study tag I consider this a self-study question so my answer is intended to help but is not complete.

There are 20 workers who earn "60 - 80" and we have no way of knowing how many of those earn more or less then "75". Thus, there is no exact answer to the question. At the other end there are 35 workers earning between "120 - 140" and there is no way for us to know how many of these earn more or less then 125.

However, we can estimate a maximum and a minimum number of workers: If we include the "60-80" and the "120-140" group then we get a maximum, if we explude it we get the minimum estimation.

Between 80 and 120 income there are $$30 + 40 = 70$$ workers.

How many are there between 60 and 140 income? How can you convert these absulute numbers to percentages?

• As there are 175 workers. Divide 70 by 175 to obtain the actual percentage of minimum number of workers who lies between 60 and 140 income i.e. 40%. Jan 11, 2021 at 11:59
• Thanks @Bernhard Jan 11, 2021 at 12:01
• Good hints. (+1). Range is between 34% and 85%. Jan 12, 2021 at 0:36

Here is a way to visualize the solution. Your answer involves intervals from 60 to 140. The only partial intervals involved are $$(60,80)$$ and $$(120,140).$$

Assuming uniform distribution within intervals. In the R program below, I have spread the required numbers of observations uniformly throughout these two intervals (vector x2). For the other intervals I have put the required numbers of observations at interval midpoints (vector x1).

Then I made a histogram (of vector x) that accounts for all $$n=175$$ observations.

x1 = rep(c(30,50,90,110,150,170, 190),
times=c(8,12,30,40,18,7,5))
x2 = c(runif(20,60,80), runif(35,120,140))
x = c(x1,x2)
cp = seq(20,200,by=20)
hist(x, br=cp, label=T, ylim=c(0,50), col="skyblue2",
main="Salaries")
abline(v=c(75,125), col="red")


So, roughly speaking, you want the observations between the two vertical red lines at $$75$$ and $$125.$$ The proportion of vectors there is obtained by taking the mean of a logical vector at the last step.

mean((x>75)&(x < 125))
[1] 0.48


On the run shown the answer is that 48% of the 175 observations fall in $$(75,125).$$ Because exact positions of values in the two intervals $$(60,80)$$ and $$(120,140)$$ are random, this proportion can vary a little from one run of the program to the next. Three additional runs gave 47.4%, 48.6% and 50.9%.

I am not sure exactly what interpolation method you are supposed to use. But it seems reasonable methods will give answers very near to 50%.

Assuming normal population. Another kind of approach uses the fact that the histogram looks roughly normal in shape. You can use formula $$\bar X \approx \frac{1}{175}\sum_{j=1}^9 f_jm_j \approx 107$$ (where $$m_j$$ and $$f_j$$ are interval midpoints and frequencies, respectively) to approximate the sample mean, which estimates the population mean $$\mu.$$ Also, $$S_x \approx \sqrt{\frac{1}{174}\sum_{j=1}^9 f_j(m_j-\bar X)^2} \approx 37,$$ which estimates the population standard deviatation $$\sigma.$$

m = seq(30, 190, by=20)
f = c(8, 12, 20, 30, 40, 35, 18, 7, 5)
a = sum(f*m)/175;  a
[1] 107.0286

v = sum(f*(m-a)^2)/174
s = sqrt(v);  s
[1] 36.89492


Then one might approximate the proportion of observations in $$(75,125)$$ by finding the probability that $$X \sim \mathsf{Norm}(107, 37)$$ has $$P(75 \le X < 125) \approx 0.49.$$ This probability can be calculated using software, as shown below in R, or by standardizing and using printed tables of the standard normal CDF.

diff(pnorm(c(75,125), 107, 37))
[1] 0.4931324

hist(x, prob=T, col="skyblue2", main="Normal Salaries")
abline(v = c(75, 125), col="red")
curve(dnorm(x, 107, 37), add=T)


• Thanks @BruceET it is very helpful solution and your estimate i.e. 48% is the correct answer. Jan 13, 2021 at 12:18