Respuesta :
Answer:
a. X[bar]= 1.52 occupants
b. S= 0.94 occupants
c. Me= 50th car
d.
C₁= 25th car
C₃= 75th car
e. P(X>1.25)= 0.30
30% of the cars had more occupants than the mean.
f. P(X> 2.19)= 0.15
15% of the cars have more than one standard deviation greater than the mean occupants.
g. P(X=2.19) = 0.85
85% of the cars have a number of occupants between one standard deviation of the mean.
Step-by-step explanation:
Hello!
There was a sample of 100 cars taken during a morning comute and the number of occupants per car vas registred:
X= number of occupats per car.
fi= absolute frequency
x: 1, 2, 3, 4, 5
fi: 70, 15, 10, 3, 2 ∑fi= n = 100
x*fi: 70, 30, 30, 12, 10 ∑x*fi= 152
x²fi: 70, 60, 90, 48, 50 ∑x²fi= 318
hi: 0.7, 0.15, 0.10, 0.03, 0.02
Hi: 0.7, 0.85, 0.95, 0.98, 1
I'll add files to the table according to the auxiliary calculations I'll need for each item.
a. To calculate the sample mean (X[bar]) you usually add every observation and divide it by the sample size, but when the data is given in a frequency table the calculation you have to make changes a little:
[tex]X[bar]= \frac{sumx*fi}{n}[/tex]
[tex]X[bar]= \frac{152}{100}[/tex]
[tex]X[bar]= 1.52[/tex]
b. The standard deviation (S) is the square root of the sample variance(S²).
So first you need to calculate de variance and then the standard deviation.
[tex]S^2= \frac{1}{n-1} [sumx^2fi-\frac{(sumx*fi)^2}{n} ][/tex]
[tex]S^2= \frac{1}{99}[318-\frac{(152)^2}{100} ][/tex]
[tex]S^2= 0.878[/tex]
[tex]S= \sqrt{S^2}[/tex]
[tex]S=\sqrt{0.878}= 0.937= 0.94[/tex]
c. The median for even sample size is calculated as:
Me= n/2= 100/2= 50
This means that the 50th car observed separates the sample in exactly 50%. The number of occupants for the 50th car was x= 1 occupant.
d.
The first quartile (C₁) separates the sample leaving 25% of the data below it and 75% of the data above it.
C₁= n/4= 100/4= 25
This means that the first quartile is the 25th observed car and the observed occupants for this car are x=1.
The third quartile (C₃) separates the sample leaving 75% of it below and 25% of it above.
C₃= n*3/4= 100*3/4= 75
The third quartile corresponds to the 75th car observed and the number of occupants in this car was x=2.
e. To know the proportions or probabilities for each value of the variable x you have to calculate their relative frequency (hi)
hi= fi/n
Note: the summary of all relative frequencies (also called relative cumulative frequencies Hi) should be 1 (max value of probabilities)
Hi= ∑hi
The proportion that has more than the mean number of occupants is the probability or proportion of cars with more than 1.25 occupants.
Symbolically:
P(X>1.25)
This expression includes x= 2 occupants, x=3 occupants, x=4 occupants and X=5 occupants.
There are two ways of obtaining this value, one is making a summary of the relative frequencies (hi) of each value of x.
P(X>1.25)= h(2)+h(3)+h(4)+h(5)= 0.15 + 0.10 + 0.03 + 0.02= 0.3
Or using the cummulative relative frequencies.
P(X>1.25)
You extract to the max probability the probabilities of the values of the variable that are not included in the expression, in this case, the only one not included is x=1, so:
P(X>1.25)= 1 - P(X ≤ 1.25)= 1 - P(X ≤ 1)= 1 - H(1)= 1 - 0.7= 0.3
30% of the cars had more occupants than the mean.
Either way, you have to reach the same result. Since for most distributions, there are tables of cumulative probabilities available and considering you usually work with variables with a larger range of definitions, instead of calculating the probability of each possible occurrence and adding them, it is much easier to use the cumulative probabilities. So I recommend you to get used to getting probabilities using this method.
f.
Now you have to calculate the proportion of cars that have more occupants than the mean + one standard deviation.
Symbolically:
P(X>1.25+0.94)= P(X> 2.19)
This expresion includes x=3, x=4 and x=5 and excludes x=1 and x=2, so using the cumulative relative prequencies you can rewrite it as:
P(X> 2.19)= 1 - P(X< 2.19)= 1 - P(X ≤ 2)= 1 - H2= 1 - 0.85= 0.15
15% of the cars have more than one standard deviation greater than the mean occupants.
g.
Now you have to calculate the proportion of cars that had within one standard deviation of the mean.
Symbolically
P(X=1.25+0.94)= P(X=2.19)
Now, this is a little more complicated because the variable of interest is a discrete variable and, to say it simply, it "jumps" from one integer to the next one, between these two values there is no change in the information.
(see attached diagram) As I've said, for this type of variable the cumulative proportion will be P(X≤2)= H(2)= 0.85 until it "jumps" to the next "step" and changes to P(X≤3)= H(3)= 0.95.
With this in mind, the proportion for P(X=2.19) equals P(X=2) = H(2) = 0.85
I hope it helps!
The sample of 100 cars is driving on the road on the freeway and during the morning the number of the occupants that are found in each car was recorded as a result of 1.2.3.4.5 and the number of cars to be 70, 15, 10, 3, 2.
- As per the question the means number of occupants included X = 1.52 occupants.
- The standard deviation of the umber of people is S= 0.94 occupants. The median number of occupants = 50th car The first and the third quartiles of numbers C₁= 25th car, and C₃= 75th car.
- The part of the cars that has more than mean no. of occupants is P (X>1.25) = 0.30 hence 30% of the cars had more occupants than the mean.
Learn more about the cars driving on a freeway.
brainly.com/question/22367242.
