Answer:
see explaination
Step-by-step explanation:
Using the formulla that
sum of terms number of terms sample mean -
Gives the sample mean as \mu=17.954
Now varaince is given by
s^2=\frac{1}{50-1}\sum_{i=1}^{49}(x_i-19.954)^2=9.97
and the standard deviation is s=\sqrt{9.97}=3.16
b) The standard error is given by
\frac{s}{\sqrt{n-1}}=\frac{3.16}{\sqrt{49}}=0.45
c) For the given data we have the least number in the sample is 12.0 and the greatest number in the sample is 24.1
Q_1=15.83, \mathrm{Median}=17.55 and Q_3=19.88
d) Since the interquartile range is Q_3-Q_1=19.88-15.83=4.05
Now the outlier is a number which is greater than 19.88+1.5(4.05)=25.96
or a number which is less than 15.83-1.5(4.05)=9.76
As there is no such number so the given sample has no outliers
Answer:
what is the rest of your qustion
Step-by-step explanation:
