Sampling distribution of sample mean

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SUMMARY

The discussion centers on calculating the probability P(X < 8) for a sample mean X derived from ten independent normal random variables, each with a mean of 10 and a standard deviation of 4. The participant correctly identifies that the expected value E(X) is 10 and the variance Var(X) is 1.6, leading to a standard deviation of approximately 1.26. The final calculation using the standard normal distribution yields P(Z < -1.58), which the participant initially misinterprets as 0 due to a mental error in using the wrong parameters for the z-table. The correct interpretation confirms that the participant's approach was fundamentally sound, despite initial doubts.

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Knowledge of calculating expected value and variance
  • Familiarity with standard normal distribution and z-scores
  • Experience using z-tables for probability calculations
NEXT STEPS
  • Review the Central Limit Theorem and its implications for sample means
  • Practice calculating probabilities using different sample sizes and distributions
  • Explore the use of statistical software for probability calculations
  • Learn about confidence intervals and their relationship to sample means
USEFUL FOR

Students in statistics, data analysts, and anyone preparing for exams involving probability and sampling distributions.

Shackman
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Homework Statement


Let X1,X2,...X10 be independent normal random variables with mean 10 and std dev 4. Let Y = X1+X2+...+X10. Let X be Y/10. What is P(X < 8)


Homework Equations


E(X) = 10 and Var(X) = 42 / 10 = 1.6
std dev of X = 1.6.5 = 1.26


The Attempt at a Solution



To convert to standard normal distribution to use z table to solve I get..

P(X-10/1.26 < (8-10)/1.26) = P(Z < -1.58) = 0

I really doubt that I have done this problem correctly because it is on a sample exam and seems too easy. Have I done anything incorrectly or am I just second guessing myself needlessly?
 
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Shackman said:
Have I done anything incorrectly

Just the typo in the last line where you say the probability is 0.

am I just second guessing myself needlessly?

Yes!
 
Ah. A silly mental error as I was using an applet to find the value for z that matched the probability but with a curve that had the mean and standard deviation of the original variable. Thanks for helping me again Billy Bob!
 

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