A sample of 100 cans of peas shown an average weight of 14 oz with a standard deviation of 0.7 Oz. Construct a 95% confidence in
terval for the mean of the population.
1 answer:
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Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
Step-by-step explanation:
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.
In this problem, we have that:
What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater