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
Step-by-step explanation:
as you know, there are 12 months per year.
also, the word "annual" means "per year" (of again 12 months).
each worker gets 7,576 per month.
that means each worker gets an annual salary of
7,576 × 12 = 90,912
the salaries/wages of all 53 workers together in a year are then
90,912 × 53 = 4,818,336
Answer:
Step-by-step explanation:
a.) The worst-case height of an AVL tree or red-black tree with 100,000 entries is 2 log 100, 000.
b.) A (2, 4) tree storing these same number of entries would have a worst-case height of log 100, 000.
c.) A red-black tree with 100,000 entries is 2 log 100, 000
d.) The worst-case height of T is 100,000.
e.) A binary search tree storing such a set would have a worst-case height of 100,000.
Answer:
7.1437 is greater
Step-by-step explanation:
Answer: 98
Step-by-step explanation:
