Find the mean, median, and mode for the data set.
A. mean = 23.6, median = 25, mode = 24
B. mean = 24, median = 23.6, mode = 25
C. mean = 24.6, median = 24, mode = 25
D. mean = 23.6, median = 24, mode = 25
9 answers
D. mean = 23.6, median = 24, mode = 25
The following is a set of sources Ivy earned on her science tests. What is the mean of her test scores?
86, 94, 88, 82, 95
A.86
B.71
C.85
D.89
86, 94, 88, 82, 95
A.86
B.71
C.85
D.89
D. 89
To find the mean, add up all the scores and divide by the total number of scores:
(86 + 94 + 88 + 82 + 95) / 5 = 445 / 5 = 89
To find the mean, add up all the scores and divide by the total number of scores:
(86 + 94 + 88 + 82 + 95) / 5 = 445 / 5 = 89
In the following list of data, find the range:
21, 28, 31, 35, 39, 43, 51, 60
A.35
B.37
C.39
D.40
21, 28, 31, 35, 39, 43, 51, 60
A.35
B.37
C.39
D.40
D. 40
To find the range, subtract the smallest value from the largest value:
60 - 21 = 39
Therefore, the range is 39.
To find the range, subtract the smallest value from the largest value:
60 - 21 = 39
Therefore, the range is 39.
Pat recorded the weights of the first 10 fish she caught and release at Mirror lake this season.
The weights were 8Ib, 6Ib, 9Ib, 6Ib, 7Ib, 5Ib, 7Ib, 6Ib, 23Ib, and 6Ib.
What is the median of the data set?
A.6
B.6.5
C.7
D.7.5
The weights were 8Ib, 6Ib, 9Ib, 6Ib, 7Ib, 5Ib, 7Ib, 6Ib, 23Ib, and 6Ib.
What is the median of the data set?
A.6
B.6.5
C.7
D.7.5
B. 6.5
To find the median, we first need to put the data in order:
5Ib, 6Ib, 6Ib, 6Ib, 7Ib, 7Ib, 8Ib, 9Ib, 23Ib
There are 10 numbers, so the median is the average of the 5th and 6th numbers:
(7Ib + 7Ib) / 2 = 6.5
Therefore, the median is 6.5.
To find the median, we first need to put the data in order:
5Ib, 6Ib, 6Ib, 6Ib, 7Ib, 7Ib, 8Ib, 9Ib, 23Ib
There are 10 numbers, so the median is the average of the 5th and 6th numbers:
(7Ib + 7Ib) / 2 = 6.5
Therefore, the median is 6.5.
Identify the outlier for the following data set:
2.4, 5.3, 3.5, 22.6, 1.8, 2.1, 4.6
A.2.4
B.2.1
C.22.6
D.3.5
2.4, 5.3, 3.5, 22.6, 1.8, 2.1, 4.6
A.2.4
B.2.1
C.22.6
D.3.5
C. 22.6
An outlier is a data point that is significantly different from other data points in a data set. One way to identify an outlier is to look for data points that are more than 1.5 times the interquartile range (IQR) away from the median.
First, we need to find the median of the data set. When the data set has an odd number of data points, the median is the middle value. In this case, the median is 3.5.
Next, we need to find the IQR. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1) of the data set. To find the quartiles, we need to put the data set in order.
1.8, 2.1, 2.4, 3.5, 4.6, 5.3, 22.6
The median of the lower half of the data set (Q1) is 2.4, and the median of the upper half of the data set (Q3) is 5.3. Therefore, the IQR is:
IQR = Q3 - Q1 = 5.3 - 2.4 = 2.9
Any data point that is more than 1.5 times the IQR away from the median is considered an outlier. In this case, the cut-off values are:
Lower bound: median - (1.5 x IQR) = 3.5 - (1.5 x 2.9) = -1.55
Upper bound: median + (1.5 x IQR) = 3.5 + (1.5 x 2.9) = 8.55
The data point 22.6 is more than 1.5 times the IQR away from the median, and it is greater than the upper bound. Therefore, 22.6 is the outlier.
An outlier is a data point that is significantly different from other data points in a data set. One way to identify an outlier is to look for data points that are more than 1.5 times the interquartile range (IQR) away from the median.
First, we need to find the median of the data set. When the data set has an odd number of data points, the median is the middle value. In this case, the median is 3.5.
Next, we need to find the IQR. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1) of the data set. To find the quartiles, we need to put the data set in order.
1.8, 2.1, 2.4, 3.5, 4.6, 5.3, 22.6
The median of the lower half of the data set (Q1) is 2.4, and the median of the upper half of the data set (Q3) is 5.3. Therefore, the IQR is:
IQR = Q3 - Q1 = 5.3 - 2.4 = 2.9
Any data point that is more than 1.5 times the IQR away from the median is considered an outlier. In this case, the cut-off values are:
Lower bound: median - (1.5 x IQR) = 3.5 - (1.5 x 2.9) = -1.55
Upper bound: median + (1.5 x IQR) = 3.5 + (1.5 x 2.9) = 8.55
The data point 22.6 is more than 1.5 times the IQR away from the median, and it is greater than the upper bound. Therefore, 22.6 is the outlier.