The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.13 mm. A random sample of 200 wafers is drawn. What is the probability that the sample mean warpage exceeds 1.305 mm?

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 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 random variable X, with mean and standard deviation , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

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

By the Central Limit Theorem

has a pvalue of 0.7054.

1 - 0.7054 = 0.2946

29.46% probability that the sample mean warpage exceeds 1.305 mm

answer: a and c are correct

step-by-step explanation:

the +5 is the vertical translation

the added negative symbol in front is the reflection over the y-axis

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: