Question

suppose a baseball player had 212 hits in a season. In the given probability distribution, the random variable X represents the number of hits the player obtained in a game x: 0, 1, 2, 3, 4, 5 p(x): 0.1147, 0.4977, 0.2822, 0.0681, 0.0131, 0.0242. Compute the standard deviation of the random variable x.

Answer 1

Are you supposed to use a computer program of a calculator for this?

Answer 2

You can. I believe there are other ways as well.

Answer 3

If you want to calculate it manually I can help. However I must log out for about an hour and will help after that if necessary :)

Answer 4

I believe I found out how to solve it. I had to find the mean and then I used a normal distribution table to find the answer. If it is not correct let me know. The answer I got is 0.9251

Answer 5

What value did you get for the mean?

Answer 6

The question does not say that the probability distribution is normal.

Answer 7

@Meg1985 Are you there?

Answer 8

oh I didn't catch that

Answer 9

my answer is wrong then

Answer 10

Just to confirm, the correct value for the mean is 1.9129. What did you get?

Answer 11

used formula (0x9)+(1x9.2)+(2X10.1)+(3x11.1)...so on

Answer 12

In this case the mean is
\[(0\times0.1147)+(1\times 0.4977)+(2\times 0.2822)+(3\times 0.0681)+(4\times 0.131)+(5\times 0.0245)=1.9129\]

Answer 13

The missing part of the summation is
\[...+(5\times 0.0245)=1.9129\]

Answer 14

The variance (the square of the standard deviation) for a random variable X taking the values
\[x _{1}, x _{2}, x _{3},.....x _{n}\]
with probabilities
\[p(x _{1}),p(x _{2}),...p(x _{n})\]
is given by
\[Var(X)=\sum_{i=1}^{n}(x _{i}-\mu)^{2}p(x _{i})\]
Do you want me to help you thru this summation?

Answer 15

thank you but I got it now. Thank you so much!

Answer 16

It was the formula that I was missing

Answer 17

You're welcome :)