5 from unit 1, lesson 9\n a. what is the five - number summary for the data?\n 0 2 2 4 5 5 5 7 11\n b. when…

5 from unit 1, lesson 9\n a. what is the five - number summary for the data?\n 0 2 2 4 5 5 5 7 11\n b. when the minimum, 0, is removed from the data set, what is the five - number summary?\n6 from unit 1, lesson 9\nwhat effect does eliminating the highest value, 180, from the data set have on the mean and median?\n25 50 50 60 70 85 85 90 90 180\n7 from unit 1, lesson 3\nthe histogram represents the distribution of the number of seconds it took for each of 50 students to find the answer to a trivia question using the internet. which interval contains the median?\na. 0 to 5 seconds\nb. 5 to 10 seconds\nc. 10 to 15 seconds\nd. 15 to 20 seconds
Answer
5a
Explanation:
Step1: Sort the data
$0,2,2,4,5,5,5,7,11$
Step2: Find minimum
The minimum is $0$.
Step3: Find first - quartile ($Q_1$)
There are $n = 9$ data points. The position of $Q_1$ is $\frac{n + 1}{4}=2.5$. So, $Q_1=\frac{2 + 2}{2}=2$.
Step4: Find median ($Q_2$)
The position of the median is $\frac{n+1}{2}=5$, so the median is $5$.
Step5: Find third - quartile ($Q_3$)
The position of $Q_3$ is $\frac{3(n + 1)}{4}=7.5$. So, $Q_3=\frac{5+7}{2}=6$.
Step6: Find maximum
The maximum is $11$.
Answer:
Minimum: $0$, $Q_1:2$, Median: $5$, $Q_3:6$, Maximum: $11$
5b
Explanation:
Step1: Remove 0 from data
The new data set is $2,2,4,5,5,5,7,11$.
Step2: Find minimum
The minimum is $2$.
Step3: Find first - quartile ($Q_1$)
There are $n = 8$ data points. The position of $Q_1$ is $\frac{n}{4}=2$. So, $Q_1 = 2$.
Step4: Find median ($Q_2$)
The position of the median is $\frac{n}{2}=4$. The median is $\frac{5 + 5}{2}=5$.
Step5: Find third - quartile ($Q_3$)
The position of $Q_3$ is $\frac{3n}{4}=6$. So, $Q_3 = 5$.
Step6: Find maximum
The maximum is $11$.
Answer:
Minimum: $2$, $Q_1:2$, Median: $5$, $Q_3:5$, Maximum: $11$
6
Explanation:
Step1: Calculate original mean
The original data set is $25,50,50,60,70,85,85,90,90,180$. The sum is $25+50+50+60+70+85+85+90+90+180 = 785$. The mean is $\frac{785}{10}=78.5$. The median: Since $n = 10$, the median is $\frac{70 + 85}{2}=77.5$.
Step2: Calculate new mean
Remove $180$. The new sum is $785-180 = 605$. The new mean is $\frac{605}{9}\approx67.22$.
Step3: Calculate new median
The new data set has $n = 9$ data points. The median is the 5 - th value, which is $70$.
Step4: Analyze changes
The mean decreases from $78.5$ to approximately $67.22$. The median decreases from $77.5$ to $70$.
Answer:
The mean and median both decrease.
7
Explanation:
Step1: Determine total number of data points
There are $n = 50$ students (data points).
Step2: Find position of median
The position of the median for $n = 50$ (even number of data - points) is between the 25 - th and 26 - th ordered data points.
Step3: Calculate cumulative frequencies
The frequencies for the intervals are:
- $0 - 5$ seconds: $22$
- $5 - 10$ seconds: $5$
- $10 - 15$ seconds: $8$
- $15 - 20$ seconds: $10$
- $20 - 25$ seconds: $4$
- $25 - 30$ seconds: $1$ The cumulative frequency up to the $0 - 5$ seconds interval is $22$. The cumulative frequency up to the $5 - 10$ seconds interval is $22 + 5=27$.
Step4: Identify interval containing median
Since the 25 - th and 26 - th data points fall within the $5 - 10$ seconds interval (because cumulative frequency up to $0 - 5$ is $22$ and up to $5 - 10$ is $27$), the interval containing the median is $5 - 10$ seconds.
Answer:
B. 5 to 10 seconds