Answer:
1. Take the Average of the distances the ball travelled each hit.
2. He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.
3. He should use Mean
4. He should use Median. It best measures skewed data
Step-by-step explanation:
THE FIRST PART.
Raul should take the average of the distances the ball travelled each hit.
This is done by summing the total distances the ball travelled each bounce, and then dividing the resulting value by the total number of times he hit the ball, which is 10.
THE SECOND PART
He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.
THE THIRD PART
He should take the mean of the distances of the ball that stayed infield.
This is the distance that occurred the most during the 9 bounces that stayed infield. The one that went outfield is makes it unfair to use any other measure of the center, taking the mean will give a value that is significantly below his efforts.
THE FOURTH PART
He should take the Median of the data, it is best for skewed data.
This is the middle value for all the distances he recorded.
Answer:
a: 28 < µ < 34
Step-by-step explanation:
We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...
We get a mean of 222/7 = 31.7143
and a sample standard deviation of: 4.3079
We can now construct our confidence interval. See the second attached photo for the construction steps.
They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)
The t-value is: 1.943
We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.
Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
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