Respuesta :
Answer:
Step-by-step explanation:
Hello!
a)
To calculate the mean of the monthly rent paid by students living off-campus you have to add al observations and divide it by the sample size:
X[bar]= ∑X/n= 21740 /30= 724.67
The mean always takes values within the range of definition of the variable, but it does not necessarily coincide with an observed value.
To calculate the Median there are two steps to follow, first you calculate its position:
For even samples PosMe= n/2= 30/2= 15 ⇒ This means the median is in the 15th position.
Now you order the sample data from lower to highest and then indentify the observation in the 15th place:
500; 560 ; 570 ; 600 ; 620 ; 620 ; 650 ; 660 ; 670 ; 690 ; 690 ; 700 ; 700 ; 710 ; 720 ; 720 ; 730 ; 730 ; 730 ; 730 ; 740 ; 740 ; 760 ; 800 ; 820 ; 840 ; 850 ; 930 ; 930 ; 1030
Me= 720
Like the mean the median always takes values within the range of definition of the variable, but it does not necesarly coincide with an observed value.
The mode is the value of the variable with more absolute frequency, in other words, the most observed value.
620 is observed 2 times
690 is observed 2 times
700 is observed 2 times
720 is observed 2 times
730 is observed 4 times
740 is observed 2 times
930 is observed 2 times
The rest of the values are observed only one time.
730 is the value with the most absolute frequency so it corresponds to the mode.
Md= 730
The standard deviation is the square root of the variance:
[tex]S^2= \frac{1}{n-1}[sumX^2-\frac{(sumX)^2}{n} ][/tex]
[tex]S^2=\frac{1}{29}[16133000-\frac{(21740)^2}{30} ] = 13060.229[/tex]
S= √S²= √13060.23= 114.28
b.
Me= 720 < X[bar]= 724.67 < Md= 730
As you can see the median, mean and mode are similar but not equal, this shows that the distribution of the data set is not exactly symmetrical but slightly skewed to the left. Since the mode is the value with more value with the more absolute frequency you can use it as a starting point to compare the three measurements of central tendency, both the mean and the median are less that it, this shows you the left-skewed tendency of the distribution.
Most correct answer: The mean or the median, because these measures are close in value, which suggests the data set is fairly symmetric.
c.
Yes (check b.)
e.
Z= [tex]\frac{800-724.67}{114.28}[/tex]= 0.659
f.
Yes there are unusual data (i.e. outliers)
An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on statistical analysis.
In this data set, there are three outliers: 500, 930 and 1030.
g.
The empirical rule states that
68% of the data is between μ-σ≤0.68≤μ+σ
95% of the data is between μ-2σ≤0.95≤μ+2σ
99% of the data is between μ-3σ≤0.99≤μ+3σ
If this is true then:
1)724.67-114.28 ≤ 0.68 ≤ 724.67+114.28
610.4 ≤ 0.68 ≤ 838.96
Under the standard normal distribution:
-Z ≤ 0.68 ≤ Z
-0.994 ≤ 0.68 ≤ 0.994
Now reversing the standardization with the sample data to check if it is equal to the empirical values:
[tex]-0.994= \frac{x-724.67}{114.28}[/tex]
x= 581.25
[tex]0.994= \frac{x-724.67}{114.28}[/tex]
x= 838.26
581.25 ≤ 0.68 ≤ 838.26
2)724.67-2*114.28 ≤ 0.95 ≤ 724.67+2*114.28
496.12 ≤ 0.95 ≤ 953.24
Under the standard normal distribution:
-Z ≤ 0.95 ≤ Z
-1.960 ≤ 0.95 ≤ 1.960
Now reversing the standardization with the sample data to check if it is equal to the empirical values:
[tex]-1.96= \frac{x-724.67}{114.28}[/tex]
x= 500.68
[tex]1.96= \frac{x-724.67}{114.28}[/tex]
x=948.66
500.68 ≤ 0.95 ≤ 948.66
3)724.67-3*114.28 ≤ 0.99 ≤ 724.67+3*114.28
381.84 ≤ 0.99 ≤ 1067.52
Under the standard normal distribution:
-Z ≤ 0.99 ≤ Z
-2.576 ≤ 0.99 ≤ 2.576
Now reversing the standardization with the sample data to check if it is equal to the empirical values:
[tex]-2.576= \frac{x-724.67}{114.28}[/tex]
x= 430.28
[tex]2.576= \frac{x-724.67}{114.28}[/tex]
x= 1019.055
430.28 ≤ 0.99 ≤ 1019.055
Comparing the empirical intervals with the data intervals, you can say that the data could be from a normal population.
I hope it helps!
The correct responses for the given data on the monthly rents paid by
students are;
- (a) Mean ≈ 724.67, Median = 720, Mode = 730
- The standard deviation is approximately 114.28
- (b) The measure of central tendency most appropriate for the data set is; The mean or the median because these measures are close in value, which suggests the data is fairly symmetric.
- (c) Yes
- (e) The z-score for a Rent value of 800 is Z ≈ 0.66
- (f) Yes
- (g) Yes
Reasons:
(a) Using MS Excel, the data are entered in a single column
To find the mean;
- An empty cell is select for the location of the result
- The Formula menu is selected followed by the More Functions Menu from which the Statistical menu is then selected
- From the options in the Statistical menu, select Average
- When the Average dialogue box comes up, select all the cells containing the 30 data entered from the question
- Select OK, then press enter. The value of the average will appear in the empty cell selected in step 1.
The Mean = 724.6667 ≈ 724.67
The median, (MEDIAN), mode, (MODE.MULT) and standard deviation, (STDEV.S) can be found using the same process above
Median (MEDIAN) = 720
Mode (MODE.MULTI) = 730
Standard deviation (STDEV.S) = 114.2814 ≈ 114.28
(b) A measure of central tendency indicates the location of the center of the data.
Given that the number of values that come before and after the mean, or
median, are almost equal, the mean or median are the most appropriate
measure of central tendency for the set.
Therefore, the correct option is; The mean or the median because these
measures are close in value, which suggests the data is fairly symmetric.
(c) The mean and the median as well as the mode are almost equal, and
therefore, Yes they agree on the center.
(e) From MS Excel, with the statistical formula STANDARDIZE, and the
above calculated mean and standard deviation, we have;
The z-score with x = 800, the formula displayed is as follows;
=STANDARDIZE(800,724.67, 114.28) = 0.65917 ≈ 0.66
The z-score for the rent value of 800 is; Z ≈ 0.66
(f) An unusual value can be taken to be an outlier.
The lower outlier are values less than = Q₁ - 1.5 × IQR
Higher outlier are values higher than= Q₃ - 1.5 × IQR
Where;
Q₁ = The first quartile
Q₃ = The third quartile
IQR = The Interquartile Range = Q₃ - Q₁
Using MS Excel function, QUARTILE.INC, we have; Q₁ = 662.5
Similarly, we have, Q₃ = 755
IQR = Q₃ - Q₁ = 755 - 662.5 = 92.5
Therefore;
Lower outlier values are lower than; 662.5 - 1.5 × 92.5 = 523.75
The lower outlier is therefore the data value of 500
Higher outlier are higher than; 775 + 1.5 × 92.5 = 913.75
The higher outliers are; 930, 930, 1030
Therefore;
Yes, The unusual values are 500, 930, 930, 1030
(g) The Empirical Rule is the 68-95-98.7 rule which is determined as
follows;
(At least)
68% of the data are within (μ ± σ)
95% of the data are within (μ ± 2·σ)
98.7% of the data are within (μ ± 3·σ)
Where;
μ = The mean ≈ 724.67
σ = The standard deviation ≈ 114.28
68% of the data are within the first standard deviation from the mean, (μ ± σ) that is the number of values within (724.67 ± 114.28)
(724.67 ± 114.28) are values between 610.39 and 838.95.
The number of data values between 610.39 and 838.95 = 22
The percentage is therefore; [tex]\dfrac{22}{30} \times 100 = 73.\overline 3 \%[/tex]
The percentage of data values between 610.39 and 838.95 = [tex]73.\overline 3 \%[/tex]
Similarly, we have;
The percentage of data that are within (μ ± 2·σ) = [tex]96.\overline 6 \%[/tex]
The percentage of data that are within (μ ± 3·σ) = 100%
Using the Empirical Rule, the data is from a normal population.
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