At the U. S open tennis championship a statistician keeps track of every serve that a player hits during the tournament. The mean serve speed was 100 miles per hour and the standard deviation of the serve speeds was 15 mph. Assume the distribution of serve speeds was mound shaped and symmetric. What percentage of the playes serves were between 115 a nd 145 mph?

Step-by-step explanation:

Both 115 and 145 mph are above the mean.  Draw a normal curve and mark these speeds.  115 mph is 1 standard deviation above the mean; 130 would be 2 standard deviations above the mean; and 145 would be 3 s. d. above it.

We need to find the area under the standard normal curve between 115 and 145.  This is equivalent to the area under the standard normal curve between z = 1 and z = 3.

I used my TI-83 Plus calculator's DISTR function "normalcdf(" to calculate this area:  normalcdf(1, 3) = 0.1573.

The area between z = 1 and z = 3 is 0.1573.  In other words, the percentage of serves that were between 115 and 145 mph was 15.73%.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."