The total from 2005 would be closer to 3,000,000 and the total from 2004 would be 3,000,000 also. So 3,000,000 + 3,000,000 = 6,000,000. I hope this helps!
Answer:
(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0
(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0
Step-by-step explanation:
(a) when i = 2, the expected number of played games will be:
E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] = 4p²-4p+2-6p²+6p = -2p²+2p+2.
If p = 1/2, then:
d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.
(b) when i = 3;
E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]
Simplification and rearrangement lead to:
E(X) = 6p⁴-12p³+3p²+3p+3
if p = 1/2, then:
d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10
Therefore, E(X) is maximized.
The dot plot that corresponds to the given data is the second dot plot
The frequency and the percent frequency distribution are shown below
<h3>
Frequency distribution</h3>
The given data is
5.6 9.9 11.3 7.5 10.0 11.9 13.2 13.8 10.0 11.9
6.5 9.0 11.2 10.9 14.6 7.2 10.0 6.0 15.8 11.3
From the question, we are to construct a dot plot
Among the given options, the dot plot that corresponds to the given data is the second dot plot
b) We are to construct a frequency distribution
The frequency distribution table is shown below
Class Frequency
6.0 - 7.9 4
8.0 - 9.9 2
10.0 - 11.9 9
12.0 - 13.9 2
14.0 - 15.9 2
Total 19
b) We are to construct a percent frequency distribution
The percent frequency distribution table is shown below
Class Percent Frequency
6.0 - 7.9 21.1%
8.0 - 9.9 10.5%
10.0 - 11.9 47.4%
12.0 - 13.9 10.5%
14.0 - 15.9 10.5%
Total 100%
The frequency and the percent frequency distribution are shown above
Learn more on Frequency distribution here: brainly.com/question/1094036
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Answer:
Step-by-step explanation:
Hello!
a)
The given information is displayed in a frequency table, since the variable of interest "height of a student" is a continuous quantitative variable the possible values of height are arranged in class intervals.
To calculate the mean for data organized in this type of table you have to use the following formula:
X[bar]= (∑x'fi)/n
Where
x' represents the class mark of each class interval and is calculated as (Upper bond + Lower bond)/2
fi represents the observed frequency for each class
n is the total of observations, you can calculate it as ∑fi
<u>Class marks:</u>
x₁'= (120+124)/2= 122
x₂'= (124+128)/2= 126
x₃'= (128+132)/2= 130
x₄'= (132+136)/2= 134
x₅'= (136+140)/2= 138
Note: all class marks are always within the bonds of its class interval, and their difference is equal to the amplitude of the intervals.
n= 7 + 8 + 13 + 9 + 3= 40
X[bar]= (∑x'fi)/n= [(x₁'*f₁)+(x₂'*f₂)+(x₃'*f₃)+(x₄'*f₄)+(x₅'*f₅)]/n) = [(122*7)+(126*8)+(130*13)+(134*9)+(138*3)]/40= 129.3
The estimated average height is 129.3cm
b)
This average value is estimated because it wasn't calculated using the exact data measured from the 40 students.
The measurements are arranged in class intervals, so you know, for example, that 7 of the students measured sized between 120 and 124 cm (and so on with the rest of the intervals), but you do not know what values those measurements and thus estimated a mean value within the interval to calculate the mean of the sample.
I hope this helps!
