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This is a question from my textbook, there is no answers at the back and I am fairly new to statistics, my answers for questions a-c are below, can anyone check if I answered them correctly?

This is a weighted average question:

Mean: E(X)= ∑〖xp(x)〗=

7(0.05)+8(0.08)+9(0.10)+10(0.14)+11(0.28)+12(0.21)+13(0.14)= 10.71

Variance:

X values: 7, 8, 9, 10, 11, 12, 13, 14

X^2 values: 49, 64, 81, 100, 121, 144, 169, 196

E(X^2) = 49(0.05)+64(0.08)+81(0.10)+100(0.14)+121(0.28)+144(0.21)+196(0.14)=121.23

Variance= E(X^2) –E(X)2=121.23-(10.71)2= 6.53

Standard deviation= √6.53 = 2.56

b) mean + one standard deviation = 10.71+2.56= 13.27

p(x>13.27)= 0 %

c) Exactly two buy size 11: (0.28)(0.28)(0.72)= 5.64%

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In your second moment, it should be $13^2=169$ instead of $14^2=196$, so your deviation and answer for (b) will change. For the last option, you haven't consider the number of different ways of choosing the two customers who bought sneakers of size $11$, i.e. the answer is going to be ${3\choose 2}$ times what you've found.

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  • $\begingroup$ Thanks for proofreading! I totally missed that $\endgroup$
    – Timmy
    Sep 29 '19 at 13:15

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