Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
2 answers:
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Answer:
a. 241; b. 206.0; c. 272.0; d. 174 to 232; other answers (see below).
Step-by-step explanation:
First, at all, we need to organize the data from the smallest to the largest value:
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
This step is important to find the median, the first quartile, and the third quartile. There are numerous methods to find the median and the other quartiles, but in this case, we are going to use a method described by Tukey, and it does not need calculators or software to estimate it.
<h3>Part a: Median</h3>In this case, we have 53 values (an odd number of values), the median is the value that has the same number of values below and above it, so what is the value that has 26 values below and above it? Well, in the organized data above this value is the 27th value, because it has 26 values above and below it:
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
The median is 241.
<h3>Part b: First Quartile</h3>For the first quartile, we need to calculate the median for the lower half of the values from the median previously obtained. Since we have an odd number of values (53), we have to include the median in this calculation. We have 27 values (including the median), so the "median" for these values is the value with 13 values below and above it.
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 <em>241</em>
Since we are asking to round the answer to one decimal place, the first quartile is 206.0
<h3>Part c: Third Quartile</h3>We use the same procedure used to find the first quartile, but in this case, using the upper half of the values.
<em>241</em> 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
So, the third quartile is 272.0
<h3>Part d: The middle 50% </h3>The second quartile is the median and "50% of the data lies below this point" (Quartile (2020), in Wikipedia). Having this information into account, 50% of the weights lies below the median 241.
Thus, the middle 50% of the weights are from 174 to 232.
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 <em>241</em>
<h3>Part e: Sample or Population 1</h3>If the population were all professional football players, the right option is:
<em>The above data would be a sample of weights because they represent a subset of the population of all football players. </em>
It represents a sample. Supposing Football Team A are all professional, they can be considered a sample from all professional football players.
<h3>Part f: Sample or Population 2</h3>If the population were Football Team A, the right option is:
<em>The data would be a population of weights because they represent all of the players on Football Team A. </em>
<h3>Part g: population mean and more</h3><h3>Part i</h3>The mean for the population of weights of Football Team A is the sum of all weights (12529 pounds) divided by the number of cases (53).
The standard deviation for the population is:
In words, we need to take either value, subtract it from the population mean, square the resulting value, sum all the values for the 53 cases (in this case, the value is 74332.67924), divide the value by 53 (1402.50338) and take the square root of it (37.45001).
Then, the population standard deviation is 37.45.
<h3>Part iii</h3>The weight that is 3 standard deviations below the mean can be obtained using the following formula:
Then, in the case of three standard deviations below the mean, z = -3.
For the player that weights 209 pounds:
For Team B, we have a <em>mean</em> and a <em>standard deviation</em> of:
And Player B weighed 229 pounds.
This value says that Player B is lighter with respect to his team than Player A, because his weight is 0.70 below the mean of Football Team B, whereas Player A has a weight that is closer to the mean of his team. So, the answer is:
<em>Player B, because he is more standard deviations away from his team's mean weight.</em>
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