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dem82 [27]
3 years ago
8

Part A: What is the mean and median amount of rainfall of this data set?

Mathematics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

Part A:  2.mean=6.42,median=7

Part B: 2.median

Part C: 2.No it is skewed left and unimodal

Step-by-step explanation:

Part A:

Finding the mean is pretty easy, you just need to divide the sum of the data values with the number of data. There are 19 data count which values spread from 2 to 10. The calculation for the mean will be:

mean = (2*1 + 3*1 + 4*2 + 5*2 + 6*3 + 7 *3 + 8 *4 + 9*2 + 10*1)/ 19

mean = 6.42105

To find the median, you have to put the data into a sequence from lowest to highest value. Then you have to find the spot where there is the same number of data above and below that spot. There are 19 data counts, so the median will be the (19+1)/2= 10th data from the highest or lowest.

The data should look like this:

2, 3, 4, 4, 5, 5, 6, 6, 6,            7,           7, 7, 8, 8, 8, 8, 9, 9, 10

The median is 7

Part B:

To choose which central tendency best used for data, you have to find how the distribution of data. Mean is easily influenced by outlier so it's not good for skewed data. Median is a better choice for skewed data.

The mode used for a categorical or discreet type of value. The data of rainfall using numerical value, so the mode is not good.

Since the data is skewed to the left (explanation on part C) then median will be best choice to represent the data.

Part C:

The normal distribution has one peak and symmetrical. Data with one peak also called unimodal. In this data, there is only one peak at 8 so it's unimodal.  

A data called symmetrical if it spread equally to the right and left. The data here is not symmetrical, it has more value on the left of 8 compared to the right of it. The lowest value on the left is 2 which is 6 value lower than the peak, while the highest value is 10 which 2 values higher than the peak. From this observation, we can say that the data is

1. skewed left

2. unimodal

Since the data skewed, its not a normal distribution.

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