Question

the given frequency distribution the students the number of passengers who board at particular day time 5-8, 8-11 ,11-14 ,14-17 ,17- 20, 20- 23 number of passengers 40 90 44 58 53 10 find the mode and median of the given data​

Answer 1

Step-by-step explanation:

The given frequency distribution represent the number of passenger who boarded a local bus during a particular day time and no. Of passenger 5-8 =40,8-11 =90,11-14 =44,14-17 =58,17-20 =53,20-23 =10. Find the mode and median of the above data

Answer 2

Median=12.19

Mode=9.56

Step-by-step explanation:

Sum of all frequencies=$N=\sum f_i=295$

N is odd therefore,

Median class=$(\frac{N+1}{2})^{th}$ observation=$\frac{295+1}{2}=148$

148 lies in interval 11-14

Median=$l+\frac{\frac{N}{2}-c.f}{f}\times h$

Where h=Size of class

N=Sum of frequencies

l=Lower limit of median class

f=Frequency of median class

c.f=Cumulative frequency of the class preceding median class

We have l=11,h=3,f=44,c.f=130

Substitute the values in the formula

Median=$11+\frac{\frac{295}{2}-130}{44}\times 3$

Median=$11+\frac{52.5}{44}=12.19$

Mode=$l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

Where l=Lower limit of modal class

$f_1=$Frequency of modal class

$f_0$ =Frequency of class preceding modal class

$f_2$=Frequency of class succeeding the modal class

h=Size of class

Modal class: The class whose frequency value is highest

Modal class=8-11

l=8,h=3,$f_1=90,f_0=40,f_2=44$

Substitute the values in the formula

Mode=$8+\frac{90-40}{2(90)-40-44}\times 3$

Mode=9.56

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