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**Shakarri** Given the independant variable with mean =ux=120 and uy=300 The St. Dev is Qx= 12, Qy=16. What is the mean and ST.D of each veriibles

a) 0.8y+20 b)2x-100 c)x+2y

There are 4 rules to remember for transferring means and standard deviations

Where A and B are constants

If the mean of x is u_{x} then the mean of Ax is Au_{x}

If the mean of y is uy then the mean of y+B is uy+B

If the St dev of x is Q_{x} then when all the values get increased by a factor of A to Ax then all the values get A times farther apart and the standard deviation gets A times larger. So the St dev of Ax is AQ_{x }If the St dev of y is Q_{y} then when all the values get increased to y+B the values just get bigger they don't get farther apart since they were all increased by the same amount and so the standard deviation doesn't change.

#2

a) Find the probability that all are right handed. 1 minus this probability is the probability that not all are right handed (one or more are left handed)

b) First find the probability that when you choose 5 people

ABCDE

the first 3 are left handed (L) and the last 2 are right handed (R)

LLLRR

Use the AND rule for this. Think of it as choosing Left and Left and Left and Right and Right

Now there are many ways you can arrange the left and right handed people such as LRLLR or RLLLR. In fact there are 5C3 ways of 3 left handers from 5 people (this is the nCr function on your calculator)

So the total probably of getting 3 left handers is [the probability of picking LLLRR] multiplied by 5C3

c) Repeat what you did in b) but find the chance that there are exactly 4 left handers, and the chance there is exactly 5, you already know from b) that the chance that there are exactly 3

The chance that there are 3 or more is [chance there are exactly 3]+[chance there are exactly 4]+[chance there are exactly 5]

d) Hint: [chance there are less than 3] + [chance there are at least 3] = 1