Answer:
a) Mean = 10, standard error = 0.7155
b) 28.77% probability of observing such a mean of something larger based on the historic mean and standard deviation.
c) 12.048 mm
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sample means with size n can be approximated to a normal distribution with mean and standard deviation
In this problem, we have that:
(a) (2 points) Assuming a near normal distribution what are the mean and standard error (standard deviation of the sample mean) of these quality checks?
Mean = 10, standard error = 0.7155
(b) (2 points) A recent sample of five wafers yielded a sample mean of 10.4 mm. Find the probability of observing such a mean of something larger based on the historic mean and standard deviation.
This is 1 subtracted by the pvalue of Z when Z = 10.4. So
By the Central Limit Theorem
has a pvalue of 0.7123.
1 - 0.7123 = 0.2877
28.77% probability of observing such a mean of something larger based on the historic mean and standard deviation.
(c) (2 points) 90% of the means taken from samples of 5 should be smaller than what value?
This is the value of X when Z has a pvalue of 0.90. So it is X when Z = 1.28.