Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 10, \sigma = 1.6, n = 5, s = \frac{1.6}{\sqrt{5}} = 0.7155[/tex]
(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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10.4 - 10}{0.7155}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex] 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.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.28 = \frac{X - 10}{1.6}[/tex]
[tex]X - 10 = 1.6*1.28[/tex]
[tex]X = 12.048[/tex]
