Hi, below is a screenshot of the question: My reasoning for the first one is tha

Question

Hi,

below is a screenshot of the question:

My reasoning for the first one is that Y is just equal to the mean of X therefore we can say that Y = E(Xi) = 5 and thus the mean would be E(E(Xi)) = 5 and the variance would be Var(5) = 0 is this valid reasoning?

I have no idea how to approach the second part of the question.

Any help would be appreciated.

Problem 5 15+5-10 points]. Suppose we want to estimate the pH of a mysterious liquid. Let the true pH be 5. We take t readings and let Xi be the value returned by the ith reading. You should assume that the readings are independent and that E(X) 5 and var(X) = 2. Let y--(X1 + X2 + … +X,)/t be the average of these readings 1) Find out the mean and variance of Y. 2) How large does t need to be such that P(4.9

Explanation / Answer

Here Mean of Y E(Y) = 5

var(Y) = Var(Xi)/ t = 2/t

(b) Now as we apply central limit theorem then we can say the distribution of Y is a normal distribution where mean of the distribution is 5 and variance = 2/t

standard deviation = (2/t)

so,

Now Pr(4.9 < Y < 5.1) = 0.99 = Pr(Y < 5.1) - Pr(Y < 4.9) = (Z2) - (Z1)

if we see z- table we get this value of probability. P = 0.99 for Z2 = + 2.575 and Z1 = - 2.575

so, here putting value

Z = 2.575 = (5.1 - 5.0)/ (2/t)

2.575 = 0.1/ (2/t)

(t/2) = 25.75

t/2 = 663.062

t = 1326.125 or say 1327

Note:

By chebyshev inequality, we can see that

1 - 1/k2 >= 0.99

1/k2 <= 0.01

k = 10

so here standard deviation of variable Y = sqrt(2/t)

so 0.1 > 10 * sqt(2/t)

sqrt(t/2) > 100

t/2 > 10000

t => 20000

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