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:
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= = 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:
x= 581.25
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:
x= 500.68
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:
x= 430.28
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!
