Respuesta :
Answer:
a) sample mean = 3.8
b) s = 0.68
c) first quartile = 3
d) median = 4
e) third quartile = 5
f) students owned at least five pairs = 35%
g) The 40th percentile = 4 pairs
h) The 90th percentile = 5 pairs
Given information:
There are total 40 students.
X = the number of pairs of sneakers owned
X Frequency
1 3
2 4
3 7
4 12
5 12
6 2
Step-by-step explanation:
3 students owned 1 pair of sneakers
4 students owned 2 pair of sneakers
and so on....
Therefore, data can also be shown like this
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, | 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6
a) Find the sample mean
sample mean = (1*3+2*4+3*7+4*12+5*12+6*2)/40
sample mean = 152/40 = 3.8
b) Find the standard deviation
[tex]s^{2}[/tex] = ∑[tex](x_{i} - x_{mean} )^{2}/n-1[/tex]
[tex]s = \sqrt{s^{2}}[/tex]
where [tex]s^{2}[/tex] is called variance and [tex]s[/tex] is the standard deviation
Putting the values in the above equation yields
s² = 18.04/39 = 0.462
s = [tex]\sqrt{0.462}[/tex] = 0.68
c) Find the first quartile
The first quartile is the middle value between starting point and median point.
The median point is 40/2 = 20th value
The first quartile is 20th - 10 = 10th value
1, 1, 1, 2, 2, 2, 2, 3, 3, 3,<= 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6
Hence the first quartile is = 3
d) Find the median
The median is 40/2 = 20th value
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, <= 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6
Hence the median is = 4
e) find the 3rd quartile
The third quartile is the middle value between median point and ending point.
The median point is 40/2 = 20th value
The third quartile is 20th + 10th = 30th value
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5,<= 5, 5, 5, 5, 5, 5, 5, 5, 6, 6
Hence the third quartile is = 5
f) What percent of the students owned at least five pairs?
students owned at least five pairs = students who owned five or more pairs
There are 12 students who owned 5 pairs = 12/40 = 0.30 = 30%
There are 2 students who owned 6 pairs = 2/40 = 0.30 = 0.05 = 5%
students owned at least five pairs = 30% + 5% = 35%
g) Find the 40th percentile
n = (P/100)*n
n = (40/100)*40
n = 16th value
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4,<= 4, 4, 4, 4, | 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6
The 16th value in the data is 4 pairs.
This means that 4 pairs of sneakers is the 20th percentile of the students.
20 percent have 4 pairs of sneakers or lower.
h) find the 90th percentile
n = (P/100)*n
n = (90/100)*40
n = 36th value
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, | 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,<= 5, 5, 6, 6
The 36th value in the data is 5 pairs.
This means that 5 pairs of sneakers is the 90th percentile of the students.
90 percent have 5 pairs of sneakers or lower.
The selected students were asked for a number of pairs of sneakers that they owned. The X was given to the number of pairs of sneakers.
As per the question answer is x = 3.8, SD 0.68, 1Q is 3, Median as 4, the Q3 as 5. The % of 5 pairs is 35 and 40% is equal to 4 parts.
- If x is the pair of the frequency 1,32,43,125,126 and 2 then sample of the mean is equal to 3.8
- The sample of the standard deviation is equal to 0.68. The first quartile will ne equal to 3, The value of the median is 4. The value of the 3rd quartile is 5
- The percent of the students that owned a least five pairs is 35% and the 40th percentile is equal to 4 pairs.
Learn more about the selected students.
brainly.com/question/14728316.
