Question

the number of adult wild cats observed in national parks is recorded: 20, 16, 11, 14, 52, 21, 19, 14, 13, 17. the park with 52 is considered an outlier. how will the mean number of adult wild cats be affected if the outlier is removed from the data set? explain your reasoning. the mean will increase because the outlier pulls the mean downward. the mean will be the same because the outlier does not affect the mean. the data must be graphed to determine how the mean will be affected. the mean will decrease because the outlier pulls the mean upward.

Explanation:

Step1: Calculate the mean with the outlier (52)

First, find the sum of all data points including 52. The data set is \(20, 16, 11, 14, 52, 21, 19, 14, 13, 17\).
Sum \(S_1 = 20 + 16 + 11 + 14 + 52 + 21 + 19 + 14 + 13 + 17\)
\(S_1 = 20+16 = 36; 36+11 = 47; 47+14 = 61; 61+52 = 113; 113+21 = 134; 134+19 = 153; 153+14 = 167; 167+13 = 180; 180+17 = 197\)
Number of data points \(n_1 = 10\)
Mean with outlier \(\bar{x}_1=\frac{S_1}{n_1}=\frac{197}{10} = 19.7\)

Step2: Calculate the mean without the outlier (52)

Now, remove 52 from the data set. The new data set is \(20, 16, 11, 14, 21, 19, 14, 13, 17\)
Sum \(S_2 = 20 + 16 + 11 + 14 + 21 + 19 + 14 + 13 + 17\)
\(20+16 = 36; 36+11 = 47; 47+14 = 61; 61+21 = 82; 82+19 = 101; 101+14 = 115; 115+13 = 128; 128+17 = 145\)
Number of data points \(n_2 = 9\)
Mean without outlier \(\bar{x}_2=\frac{S_2}{n_2}=\frac{145}{9}\approx16.11\)? Wait, no, wait, I made a mistake. Wait, 20+16=36, +11=47, +14=61, +21=82, +19=101, +14=115, +13=128, +17=145? Wait, no, original data with 52: let's recalculate the sum. Wait 20+16=36, +11=47, +14=61, +52=113, +21=134, +19=153, +14=167, +13=180, +17=197. Correct. Now without 52: 20+16=36, +11=47, +14=61, +21=82, +19=101, +14=115, +13=128, +17=145? Wait, no, there are 9 numbers? Wait original data: 20,16,11,14,52,21,19,14,13,17. That's 10 numbers. Remove 52, so 9 numbers: 20,16,11,14,21,19,14,13,17. Let's sum again: 20+16=36; 36+11=47; 47+14=61; 61+21=82; 82+19=101; 101+14=115; 115+13=128; 128+17=145. Wait, but 145 divided by 9 is approximately 16.11? But wait, the outlier is 52, which is much larger than the other values. Wait, no, wait, I think I messed up the sum. Wait 20+16=36, +11=47, +14=61, +21=82, +19=101, +14=115, +13=128, +17=145. Yes. But the mean with outlier was 19.7, and without is ~16.11? Wait, no, that can't be. Wait, no, wait, 52 is an outlier (much larger), so removing it should decrease the mean? Wait, no, wait, 19.7 is higher than 16.11? Wait, no, 19.7 is higher, so when we remove the outlier (52, which is high), the mean should decrease? Wait, but let's check the options. The options are:

1. The mean will increase because the outlier pulls the mean downward. (Incorrect, outlier is high, so it pulls mean upward)
2. The mean will be the same because the outlier does not affect the mean. (Incorrect)
3. The data must be graphed to determine how the mean will be affected. (Incorrect, we can calculate)
4. The mean will decrease because the outlier pulls the mean upward. (Correct, because 52 is an outlier (higher than most data), so it increases the mean. Removing it will make the mean lower, i.e., decrease the mean because the outlier was pulling it upward)

Wait, my calculation earlier: mean with outlier: 197/10=19.7. Mean without: 145/9≈16.11. Wait, 19.7 is higher than 16.11, so when we remove the outlier (52), the mean decreases. Because the outlier (52) is a large value, it was pulling the mean upward. So removing it, the mean decreases. So the correct option is the fourth one: "The mean will decrease because the outlier pulls the mean upward."

Answer:

The mean will decrease because the outlier pulls the mean upward. (The option corresponding to this statement, likely the last option in the given choices)