Question

The table below shows the time intervals (hours) it takes people to arrive at a counter at a bus terminal Time (hrs) Number of people 0-0.25 0.25-0.50 0.50-0.75 0.75-1.00 1.00-1.25 1.25-1.50 38 67 50 36 30 29 Use this to answer questions 5 to 8. 5. What is the modal arrival time, correct to 2 decimal places? Answer: hours 6. Find the mean time of arrival in minutes, correct to 3 significant figures? Answer: 7. What is the standard deviation of the data distribution, correct to 2 decimal place? Answer: hours 8. The median time of arrival approximated to 2 decimal places is ... Answer: hours

Answer 1

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