Mathematics, 18.01.2020 05:31 kailahgranger

Given the following cumulative frequency table, what is the mean of the distribution?

value cumulative frequency
1 0.15
3 0.45
5 0.60
7 0.85
9 1.00

Answers: 1

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Answer from: havenlynn27

The mean of the distribution is 4.9.

The calculation is as follows:

Here we have to determine the frequency from the cumulative frequency.

The Starting number is similar.

Now

To determine the next number subtract the previous number from that number.

i.e first number = 0.15

Second number = 0.45 - 0.15 = 0.3

Third number = 0.60 - 0.45 = 0.15

Fourth number = 0.85 - 0.60 = 0.25

Fifth number = 1.00 - 0.85 = 0.15

Now

Value Frequency Value × Frequency

1 0.15 0.15

3 0.3 0.9

5 0.15 0.75

7 0.25 1.75

9 0.15 1.35

Number of frequency = 1

Now

Total value should be

= 0.15 + 0.9 + 0.75 + 1.75 + 1.35

= 4.9

So, the mean is 4.9

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Answer from: acrooks86

4.9

Step-by-step explanation:

Let us first find the frequency from the cumulative frequency.

Starting number is same.

To find the next number subtract the previous number from that number.

i.e first number = 0.15

Second number = 0.45 – 0.15 = 0.3

Third number = 0.60 – 0.45 = 0.15

Fourth number = 0.85 – 0.60 = 0.25

Fifth number = 1.00 – 0.85 = 0.15

Value Frequency Value × Frequency

1 0.15 0.15

3 0.3 0.9

5 0.15 0.75

7 0.25 1.75

9 0.15 1.35

Number of frequency = 1

Total value = 0.15 + 0.9 + 0.75 + 1.75 + 1.35 = 4.9

Mean = $\frac{\text { Total value }}{\text { Number of frequency }}=\frac{4.9}{1}$ = 4.9

Answer from: Quest

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

Answer from: Quest

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

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Given the following cumulative frequency table, what is the mean of the distribution?

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