Answer:
Step-by-step explanation:
Given that a professor sets a standard examination at the end of each semester for all sections of a course. The variance of the scores on this test is typically very close to 300.
(Two tailed test for variance )
Sample variance =480
We can use chi square test for testing of hypothesis
Test statistic =
p value = 0.0100
Since p <0.05 our significance level, we reject H0.
The sample variance cannot be claimed as equal to 300.
Answer:
0.2611 = 26.11% probability that exactly 2 calculators are defective.
Step-by-step explanation:
For each calculator, there are only two possible outcomes. Either it is defective, or it is not. The probability of a calculator being defective is independent of any other calculator, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
5% of calculators coming out of the production lines have a defect.
This means that
Fifty calculators are randomly selected from the production line and tested for defects.
This means that
What is the probability that exactly 2 calculators are defective?
This is P(X = 2). So
0.2611 = 26.11% probability that exactly 2 calculators are defective.