How does the speed of a runner vary over the course of a marathon (a distance of 42.195 km)? Consider determining both the time to run the first 5 km and the time to run between the 35-km and 40-km points, and then subtracting the former time from the latter time. A positive value of this difference corresponds to a runner slowing down toward the end of the race. The accompanying histogram is based on times of runners who participated in several different Japanese marathons.†

The observed typical difference value of mode is 100 seconds approximately.

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

Step-by-step explanation:

Fundamentals

Consider that there are x favorable cases to an event E, out of a total of n cases. Then, the probability of that event is written as:

P(E)=  

Total number of cases /Number of favorable cases =  x/n

A histogram is constructed for continuous data, which is divided into classes called bins. The shape of the distribution can be determined from the histogram.

step 1

The provided histogram indicates time difference on xx -axis and frequency of runners on yy -axis. To determine the typical difference value, identify the peaks of the graph.

The histogram is skewed towards right side. The graph indicates that there are few outliers around 700 seconds. For a typical difference value, the value of mode is considered.

The graph indicates that the value of mode is 100 seconds approximately.

The observed typical difference value of mode is 100 seconds approximately.

Explanation

From the histogram, the typical difference value is obtained on the basis of guessing the value of mode which has been approximated to be around 100 seconds.

Step 2

From the histogram, it can be estimated that there are around 10 runners which has negative difference which means approxi8matley 10 runners ran the late distance more quickly than the early distance,

The approximate sample size can be calculated as:

Sample size=90+190+180+160+120+80+60+40+30+20

Thus, the proportion of runners is obtained as:

\begin{array}{c}\\p = \frac{{\left( \begin{array}{l}\\{\rm{Number of runners who has }}{\rm{negative time difference}}\\\end{array} \right)}}{{{\rm{Total sample size = \frac{{10}}{{970}} = 0.01\\\end{array}  

p=   Number of runners who has

    negative time difference  

      Total sample size

=10/970

=0.001

The proportion of runners ran the late distance more quickly than the early distance is approximately 1%

EXPLANATION

The obtained proportion is 0.01. It indicates that there are approximately 1% of the runners who ran late distance more quickly than the early distance is very few.

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.

where are the choices?