- #1
giddy
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So from what I understand, the central limit theorem allows us to calculate the probability of the mean of a number of independent observations of the same variable.
I probably have not understood something because I can't really solve any of the problems just based on the formula give.
An unbiased dice(6 sides) is thrown once. From its distribution its mean is 3.5 and vriable is 35/12. The same dice is thrown 70 times. 1. Find the probability that the mean score is less than 3.3. 2. Find the probability that the total score exceeds 260.
[tex]
X \sim N(\mu,\frac{\sigma^2}{n})
[/tex]
[tex]
= X \sim N(3.5,\frac{35/12}{70})[/tex]
So I have to find P(X<3.3) standardize X to Z
[tex]
P(Z < \frac{3.5-3.3}{\sqrt{0.0416}})
= P(Z < 0.9806)
=\phi(0.9806) = 0.8365
[/tex]
And the answer is supposed to be 0.155!
I'm not sure at all how to approach the second part of the problem!?
I probably have not understood something because I can't really solve any of the problems just based on the formula give.
Homework Statement
An unbiased dice(6 sides) is thrown once. From its distribution its mean is 3.5 and vriable is 35/12. The same dice is thrown 70 times. 1. Find the probability that the mean score is less than 3.3. 2. Find the probability that the total score exceeds 260.
Homework Equations
[tex]
X \sim N(\mu,\frac{\sigma^2}{n})
[/tex]
The Attempt at a Solution
[tex]
= X \sim N(3.5,\frac{35/12}{70})[/tex]
So I have to find P(X<3.3) standardize X to Z
[tex]
P(Z < \frac{3.5-3.3}{\sqrt{0.0416}})
= P(Z < 0.9806)
=\phi(0.9806) = 0.8365
[/tex]
And the answer is supposed to be 0.155!
I'm not sure at all how to approach the second part of the problem!?
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