Stats question , have work and answer shown

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SUMMARY

The discussion centers on calculating the glucose level threshold (L) for Sheila's test results, given a normal distribution with a mean (μ) of 125 mg/dl and a standard deviation (σ) of 10 mg/dl. The user initially calculated L using the formula x = μ + zσ, resulting in a value of 130.199 mg/dl. However, the calculation must account for the sample size (n=4), which affects the standard deviation of the sample mean, leading to a new standard deviation of σ/√n. This adjustment is crucial for determining the correct threshold for the probability of 0.05.

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Homework Statement



i didnt know which forum stats is posted in so i posted it here, but the problem says the q is , Sheilas measure glucose level one hour after a sugary drink varies according to the normal distribution (u= 125 mg/dl) and r=10 mg/dl. What is the level L such that there is a probability only 0.05 that the mean glucose level of 4 test results falls above L?

Homework Equations



x= u(mew) + z*r

The Attempt at a Solution


I did a backwards normal calculation of x= 125 + .5199 (from table A probs) * 10 = 130.199

i think its right but i dunno, does the 4 play a part in my equation or is this the wrong way to do it? ANY information would be helpful
 
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Yes, it does. The average of n samples from a population with mean [itex]mu[/itex] and standard deviation [itex]\sigma[/itex] is normally distributed with mean [itex]mu[/itex] and standard deviation [itex]\sigma\sqrt{n}[/itex]. Notice that means that the larger sample you have the smaller its standard deviation is and the more "accurate" it is.
 

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