Answer:
a)= 2
b) 6.324
c) P= 0.1217
Step-by-step explanation:
a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)
b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by
σ_X`1- X`2 = √σ²/n1 +σ²/n2
Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2
so
σ_X`1- X`2 =√20 +20 = 6.324
if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.
Where as Z is normally distributed with mean zero and unit variance.
If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then
Z= 0-2/ 6.342= -0.31625
P(-0.31625<z<0)= 0.1217
The probability would be 0.1217
Use a number line if you need 4/10 - 6/10 = -2/10a simpler way is to do the opposite 6/10 - 4/10 = 2/10 then the answer would be the opposite as well. so, -2/10
There are 30 days in April.
If 18 days had rain, it means 12 didn't.
18/30=x/100
Make sure both fractions can be made into fractions with a denominator of 100:
180/300=x/100
Divide 180/300 by 3:
60/100=x/100
This shows that the 18 days that did rain take up 60% of the whole month.
So then the remaining 12 days that didn't have rain must take up 40% of the month.
Solution:
40% of the days did not have rain during the month of April
The answer is 443 m^2
I explained it in the picture.
Hope it helps!
