Answer:
Explanation:
5.) here no. of people represent frequencies, so modal group (the group with the highest frequency) is 0.25-0.50.
Estimated Mode = L + (( fm − fm-1) / ( (fm − fm-1) + (fm − fm+1) ) ) × w
where,
L is the lower class boundary of the modal group = 0.25
fm-1 is the frequency of the group before the modal group = 38
fm is the frequency of the modal group = 67
fm+1 is the frequency of the group after the modal group = 50
w is the group width = 0.25
mode= 0.25 + ((67-38)/((67-38)+(67-50)))* 0.25
= 0.25 + (29/ (29+17))*0.25
= 0.25 + 0.63*0.25
= 0.41
6) mean= total(fx) / total(f)
= 166.25/250
= 0.665
7) standard deviation = sqaure root (( total(fx2) - (total(f)* mean2)) / (total(f)-1))
= sqaure root (( 149.906 - 250* 0.6652)/ 249 )
= square root ( (149.906 - 110.556) /249)
= sqaure root (0.158)
= 0.397
8) The median is the middle value, which in our case is the 125 (250/2) , which is in the 0.5 - 0.75 group.
Estimated Median = L + ( ((n/2) − B)/G) × w
where:
L is the lower class boundary of the group containing the median = 0.5
n is the total number of values = 250
B is the cumulative frequency of the groups before the median group = 105
G is the frequency of the median group = 50
w is the group width = 0.25
median = 0.5 + (((250/2)-105)/50)*0.25
= 0.5 + ((125-105)/50)*0.25
= 0.5 + (20/50)*0.25
= 0.6