Question

3. Find the mean (p). (Round your answer to 3 decimal places) (2 points) the mean=8.177 6 7 8 9 10 P(x) 0.236 0.063 0.214 0.262 0.225 4. Find the standard deviation (C) (Round your answer to 4 decimal places.) (3 points) the standard deviation=1.462 Suppose that samples of 100 are generated from the probability distribution and the mean of each sample is recorded. (2 points) 5. Find the mean of the sample means. (3 points) 6. Find the standard deviation of the sample means

Answer 1

Selection ; - 030 given x1 PKD 0.236 7 0.063 6 8 0.214 9 0.262 0.225 10 u. mean the (3) Find u= E(X)= Exp(x) +8x0,214 6X 0.236 + 7X01063 +90.262 + 10 X 0.225 + 2.359 1.416 + 0.441 + 1.712 EW- + 2.250 E(x) 8.177 Now E(x3 = Expla) 36 x 0.236 + 49x0.063 +64 X0.214 10 X 0.225 +81 x 0.262 8.496 + 3.087 + 13.696 21.222 + 22.500 t E (2) - 69.00)

Now Find the standard deviation € 673) - LEKIJ? 69.001 (8.177)? 69.00/ 66.863 1.462 Now sample Size (n) 100 6 Find the mean of the sample means. ie. EX) El too & Xu] loo 100 E (XL) Too & Too X 100 X 8.177 Elã) - 8.177 the Find Standard deviation of the Sample means. ie. o (8) oliva & x.) too anxi) too & ore (X) 2(X) Joo to o(X) = it x 1.462 = 0.1462 Ang

Ot@:- given Х PK) o o. 205 2 0.166 P(2) 0:191 ( 0.135 4 5 0.169 value of a what is the P (2) ? we know that E P() = 1 > 0.205 + 0.166 + P(2) + 0.191 +0.135 +0.169 = 1 = 3 0.866 + P(2) P(2) 1 -0.866 P 2) = 0.134 what is mean ш) b u= EX) Exp66) Ox0.205 + 1 X 0.166 + 2 x 0.134 + 3x0.191 + 4 X0.135 + 5x0.169 = O + 0.166 + 0.268 +0.573 + 0.540 + 0.845 u= 2.392 Ans