Question

60 randomly selected students were asked how many siblings were in their family. Let X = the number of pairs of siblings in the student's family.
The results are as follows:

Siblings Frequency
1 13
2 22
3 15
4 6
5 3
6 0
7 1

Round your answers to two decimal places.
1. The mean is ___.
2. The median is ___.
3. The sample standard deviation is ___.
4. The first quartile is ___.
5. The third quartile is ___.

Answer 1

Answer:

1. mean=2.47

2. median=2

3. Sample Standard deviation=1.24

4. First quartile=2

5. Third quartile=3

Step-by-step explanation:

The necessary calculations for finding mean,median,standard deviation, 1st quartile and 3rd quartile are

sumf=13+22+15+6+3+0+1=60

sumfx=13*1+22*2+15*3+6*4+3*5+0*6+1*7=148

sumfx²=13*1²+22*2²+15*3²+6*4²+3*5²+0*6²+1*7²=456

x   cumulative frequency

1    13

1    23+22=35

3   35+15=50

4    50+6=56

5    56+3=59

6    59+0=59

7     59+1=60

1.

mean=sum(fx)/sumf

Here x is number of siblings and f is frequency

mean=148/60=2.4667

So, after rounding to two decimal places our mean is 2.47.

2.

Median=value of (n/2)th observation

Median=value of (60/2)th observation

Median=value of 30th observation

The cumulative frequency shows that the 30th observation corresponds to x=2 sibling. So, the median is 2.

3.

$s=\sqrt{\frac{sumfx^{2} -\frac{(sumfx)^{2} }{n} }{n-1} }$

$s=\sqrt{\frac{456 -\frac{(148)^{2} }{60} }{59} }$

$s=\sqrt{\frac{456 -365.0667 }{59} }}$

$s=\sqrt{1.5412 }$

s=1.2415=1.24

4.

1st quartile=Q1=((n+1)/4)th values of observation

Q1=((61)/4)th values of observation

Q1=(15.25)th values of observation

15.25th value also corresponds to x=2 siblings so,

Q1=2.

5.

3rd quartile=Q3=(3(n+1)/4)th values of observation

Q3=(3(61)/4)th values of observation

Q3=(45.75)th values of observation

45.75th value  corresponds to x=3 siblings so,

Q3=3