What is 4 divided by 3/5
1 answer:
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Answer:
1). Z score = 1.464733
Mean = 38.87917
Standard deviation = 3.609405
2). Quartiles:
Q1 = 37.025
Q2 = 38.450
Q3 = 40.800
3). IQR = 3.775
4).
CI (lower fence) = 37.4351
CI (upper fence) = 40.32323
5). There is outlier in the data set. Please see attached box plot for evidence.
Step-by-step explanation:
1). By Z score, we mean:
Z = ,
where:
x-bar ==> mean(g) = 38.87917
= 37.80
standard deviation = 3.609405
sample size (n) = 24.
2). By quantile, we mean:
Q = L + (i*(n/4) - Cf)*c
Where L is the lower class limit of the quartile class
Cf is the cumulative frequency before the quartile class
c is the class size.
3) . IQR = Q3 - Q1
4). CI = ,
Where ==> 1.96
In order to replicate and obtain the result, please use the R code below:
g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,
41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)
boxplot(g)
Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))
mean(g)
quantile(g)
IQR(g)
CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))
CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))
Given:
The data set is
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35
To find:
Effect on mean, median, range after removing the outlier, 35.
Solution:
We have, the data set
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35
Mean of the data set is 15.12.
because n=17, which is odd.
Median is 14.
Range is 25.
After removing the outlier, 35, the data set is
10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17
Mean of the new data set is 13.875, which is less than 15.12.
because n=16, which is even.
New median is 14, which is same as original median.
So, the new range is 7 which is less than 25.
Therefore, the mean of the data set will decrease, the median of the data set will not change, the Range of the data set will decrease.