Assume that the points scored by the winning teams for all NCAA games follow a bell-shaped distribution. Using the mean of 76.5 and the standard deviation of 7, estimate the percentage of all NCAA games in which the winning team scores 84 or more points. 90 points. The winning scores are as follows: 90, 85, 75, 78,71, 65, 72, 76, 77, 76. The answers are 16% and 2.5% I have tried to solve the problem by determining the z score (84-76.5/7 = 1.072) and then use chebyshev's theorem (1 - 1 / 1.072^2 = .13) this is not quite right....what am I missing? I see that only 2 teams scored 84 points or higher and only 1 scored 90 or higher. I would follow the same equations as above for 90.
Look up Z score in table in back of statistics text labeled something like "areas under normal distribution" for calculated Z scores. The colummn for the smaller area should give the proportion you desire. Convert to precentage.
I hope this helps. Thanks for asking.
To estimate the percentage of NCAA games in which the winning team scores 84 or more points, you can use the concept of z-scores and the standard normal distribution.
To find the z-score for a winning score of 84, you can use the formula:
z = (x - μ) / σ
where:
x = individual score (84 in this case)
μ = mean (76.5)
σ = standard deviation (7)
So, z = (84 - 76.5) / 7 = 1.0714 (rounded to four decimal places)
Then, you can use a standard normal distribution table or calculator to find the corresponding area under the curve to the right of this z-value. This represents the percentage of games where the winning team scored 84 or more points.
Using a standard normal distribution table, the area to the right of z = 1.07 is approximately 0.1423. This means that about 14.23% of games would have a winning team score of 84 or more points.
For the percentage of games in which the winning team score 90 or more points, you can follow the same steps:
z = (90 - 76.5) / 7 = 1.9286 (rounded to four decimal places)
Using a standard normal distribution table, the area to the right of z = 1.93 is approximately 0.0276. This means that about 2.76% of games would have a winning team score of 90 or more points.
To estimate the percentage of NCAA games in which the winning team scores 84 or more points, you can use the concept of z-scores and the standard normal distribution.
First, let's calculate the z-score for the value 84 using the formula:
z = (X - μ) / σ
where X is the value we want to find the probability for (84), μ is the mean (76.5), and σ is the standard deviation (7).
Plugging in the values, we get:
z = (84 - 76.5) / 7 = 1.071
Now, to find the probability associated with this z-score, we can look up the corresponding value in a standard normal distribution table. These tables typically provide probabilities for values up to certain z-scores.
However, if we use the Chebyshev's theorem, it only provides a lower bound on the probability. Chebyshev's theorem states that no matter the shape of the distribution, at least (1 - 1 / k^2) of the data falls within k standard deviations from the mean. In this case, k is equal to the z-score (1.071).
So, using Chebyshev's theorem, we can estimate that at least (1 - 1 / 1.071^2) = (1 - 1 / 1.1474) = 0.1276 (or 12.76%) of all NCAA games will have winning teams scoring 84 or more points.
However, for a more accurate estimate, we can use the cumulative distribution function (CDF) of the standard normal distribution to find the exact probability. This requires either using a standard normal distribution table or a statistical software/tool.
Using the standard normal distribution table or a tool, you can find that the probability corresponding to a z-score of 1.071 is approximately 0.8577 (or 85.77%).
So, the estimated percentage of NCAA games in which the winning team scores 84 or more points is approximately 85.77%.
To estimate the percentage of games in which the winning team scores 90 or more points, you would follow the same steps. Calculate the z-score for 90, look up the corresponding probability in the standard normal distribution table, and convert it to a percentage.
Using the same formula as before, the z-score for 90 would be:
z = (90 - 76.5) / 7 = 1.93
Looking up the probability associated with a z-score of 1.93 in the standard normal distribution table, you will find that it is approximately 0.9744 (or 97.44%).
So, the estimated percentage of NCAA games in which the winning team scores 90 or more points is approximately 97.44%.
Please note that these percentages are estimates based on the information provided and assumptions about the distribution.