What is the GCF of 44j5k4 and 121j2k6? 4j3k2 4j2k4 11j3k2 11j2k4
2 answers:
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Answer:
0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
Normal probability distribution.
Problems of normally distributed samples can be 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.
In this problem, we have that:
Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
This probability is the pvalue of Z when . So
has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.