We are to find t here such that:
P( M <= t) <= 0.9
The CDF for M is obtained here as:
P(M <= m) = Probability that the maximum value is less than or equal to m
P(M <= m) = 1 - P( M > m)
P(M <= m) = 1 - (1 - Probability that all the Xi are less than m)
We are given here for each i = 1, 2, 3, .. 10 that:
Therefore, the CDF for M here is obtained as:
Therefore, we are to find t here such that:
From standard normal tables, we have here:
P(Z < 2.309) = 0.9895
Therefore t = 2.309 is the required 90th percentile value of M here.
Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let...
Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.
Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.
Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.
Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.
Question 4 10 pts Suppose X1, X2, ..., X10 are independent normal random variables with mean O and variance 1. Let M = max{X1, X2,..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M. Upload Choose a File
I got 2.308 for this question Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.
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