Question

Suppose X∼N(50,2.5) represents a population, if X 1​ ,X 2 ,…,X n

​ is a random sample from this population, find n such that P(X 1 +X 2​ +⋯+X n​ >2000)=0.95

Answer 1

To find n such that P(X1 + X2 + ... + Xn > 2000) = 0.95, convert the population values to z-scores, find the corresponding percentile, and solve for n using the equation (n * µ) = (n * 50) = 54.536. n ≈ 1.091.

Step-by-step explanation:

To find the value of n such that P(X1 + X2 + ... + Xn > 2000) = 0.95, we need to use the properties of a normal distribution.

First, we convert the given population values to z-scores by using the formula z = (X - µ) / σ.

For X = 2000, mean µ = 50, and standard deviation σ = √2.5.

By substituting these values into the formula, we get z = (2000 - 50) / √(2.5) = 54.536.

Now, we can find the corresponding percentile by using the invNorm() function on a calculator or statistical software.

The value is invNorm(0.95) = 1.645.

Since we are summing n samples, the sample mean is n times the population mean. So, we have (n * µ) = (n * 50) = 54.536.

By solving this equation, we find n ≈ 1.091.