Respuesta :
Answer:
29.46% probability that the sample mean warpage exceeds 1.305 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 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 = 1.3, \sigma = 0.13, n = 200, s = \frac{0.13}{\sqrt{200}} = 0.0092[/tex]
A random sample of 200 wafers is drawn. What is the probability that the sample mean warpage exceeds 1.305 mm
This is 1 subtracted by the pvalue of Z when X = 1.305. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1.305 - 1.3}{0.0092}[/tex]
[tex]Z = 0.54[/tex]
[tex]Z = 0.54[/tex] has a pvalue of 0.7054.
1 - 0.7054 = 0.2946
29.46% probability that the sample mean warpage exceeds 1.305 mm
