Two angles are supplementary. One angle is 50° more than the other. Find the measure of the two angles.
2 answers:
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Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.
Using the binomial distribution, we want all three to be defective.
Using the binomial distribution, we want all three to be defective.
Answer:
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
Step-by-step explanation:
To solve this question, we need 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 sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.
Lesser than 207.9.
pvalue of Z when X = 207.9. So
By the Central Limit Theorem
has a pvalue of 0.2743
2*0.2743 = 0.5486
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches