Question

Q4- The shoe sizes of 8 students were:
8, 6, 4, 6, 6, 7, 9, 10
a) What is the modal size?
b) What is the median size?
c) Which of the above two results will be more​

Answer 1

Step-by-step explanation:

Given:-

The shoe sizes of 8 students were:

The shoe sizes of 8 students were:8, 6, 4, 6, 6, 7, 9, 10

To find:-

a) What is the modal size?

a) What is the modal size?b) What is the median size?

a) What is the modal size?b) What is the median size?c) Which of the above two results will be more.

Solution:-

Given observations are 8, 6, 4, 6, 6, 7, 9, 10

On writting them into ascending order

=>4,6,6,6,7,8,9,10

a)Most occuring observation =6

6 occurs 3 times

Mode=6

Mode:-

The most frequent occuring observation of the given data is the mode of the given data.

b) No. of observations=8

It is an even number.

We know that

If number of observations"n" is an even number then the median is the average of (n/2) th observation and (n/2) +1 th observation

=>n/2=8/2=4

(n/2)+1=4+1=5

4th observation=6

5th observation=7

Average of 6 and 7=(6+7)/2

=>13/2

=>6.5

Median=6.5

c)Median=6.5

Mode=6

6.5>6

Median is more than Mode of the data.

Answer:-

a)Modal size=6

b)Median size=6.5

c)Median is more than Mode of the data.

Answer 2

Answer:

.

Modal size = 6

.

Median = 6.5

.

Step-by-step explanation:

$ascending \: order \: of \: this \: data \\ \\ 4, 6, 6,6, 7,8 ,9, 10 \\ \\ ans > 1 \: \: \: modal \: size = 6 \\ \\ ans > 2 \: \: \: \: median = \frac{ {( \frac{n}{2}) }^{th \: \: term} + { (\frac{n}{2} + 1 )}^{th \: \: term} }{2} \\ \\ median = \frac{ { (\frac{8}{2}) }^{th \: \: term} + { (\frac{8}{2} + 1)}^{th \: \: term} }{2} \\ \\ median = \frac{ {4}^{th \: \: term } + {5}^{th \: \: term} }{2} \\ \\ median = \frac{6 + 7}{2} \\ \\ median = \frac{13}{2} = 6.5 \\ \\ ans > 3 \: \: \: median \: is \: more \\ \\ this \: is \: ur \: required \: answer \: :):):):):)$