Question

How do you find the first and third quartile of this dot plot?

Answer 1

in order to find the first quartile, you would first need to find the second quartile, or the median, and then in order to find the 1st quartile you would get the numbers on the left half of the median and find the median of those numbers, you would then do the same for the right half. Hope this helps.

Answer 2

You can make a frequency table along with cumulative frequency and then can see where does the 25% and 75% of the population comes in total.

The first and third quartile of the given dot plot is given as:

First quartile is at 14.5 minutes

Third quartile is at 57.5 minutes

What are the first, second and third quartile in a given data?

In a given data which is sorted either increasingly or decreasingly, the value before which 25% of the population resides and after which 75% of the population resides is called first quartile.

The value which has 50% of the population on left and 50% of the data on the right side is called median or second quartile of the data.

The value to which the left side is containing 7% of the data and the right side is containing 25% of the data is called third quartile of the data.

These three quartile points divide the whole data in four equal parts called quarters (a quarter is one-fourth of the total part).

How to find first and third quartile of the given data/

For that, firstly we will have to know how many of the times, a particular delay interval is observed. Then we will take value before which 25% of the delay observations lie as the first quartile, and the value before which 75% of the observation lies as the third quartile.

Using given graph, making the frequency and cumulative frequency table

$\begin{array}{ccc}\text{Delay(in minutes)}&\text{Freq.}&\text{Cumul. Freq. (CF)}\\0&2&2\\5&1&3\\10&3&6\\15&10&16\\20&7&23\\25&4&27\\30&5&32\\35&4&36\\40&3&39\\45&2&41\\50&1&42\\55&2&44\\60&2&46\\65&2&48\\70&1&49\\75&1&50\\80&1&51\\85&1&52\\90&1&53\\95&2&55\\100&17&56\\105&0&56\\110&1&57\\115&1&58\\120&2&60\\\end{array}$

There are total 60 observations.

• Taking 25% of 60

$25 \times 60/100 = 15$

Since at delay of 10 minutes, the total number of observations is 6, and at delay of 15 minutes, total number of observations is 16, thus, 15 lies in middle of those two delay time. Thus, the first quartile is between 10 and 15 minutes.

The increment was of 10 observations in 5 minute increment. Thus, 2 observation per 1 minute on average. We need 15 obs. -6 obs. = 9 observations. After 4 minute increment, and a half more increment, 9 observations will lie averagely(since 1 minute = 2 observations increment, thus 4.5 minutes will lead to 9 observation increment), thus:

First quartile will be such that 6 obs + 9 obs = 15 obs should be there(25%)

Since 6 obs will be covered on left of 10 minutes and 9 more obs will covered by 4.5 minutes,

thus 10 + 4.5 minutes will have approx 25% observations(frequency is number of observations) on left.

Thus, first quartile is at 14.5 minutes.

• Taking 75% of 60 = 100% - 25% of 60 = 60 - 15 = 45

Since 45 observations lie between 55 and 60 minutes delay, and as at 55 minutes , we had 44 observations and at 60 minutes  we had 46 observations, thus, as 45 observations is mean of those two, thus, 45 observations will lie in mean of 55 and 60 which is $\dfrac{55+60}{2} = 57.5$
Thus,

Third quartile = 57.5 minutes

Thus, the first and third quartile of the given dot plot lies on 14.5 minutes and 57.5 minutes respectively.

Learn more about quartiles here:

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