Bepotastine besilate (BEPREVE) ophthalmic solution contains 1.5% w/v of the therapeutic agent. Express this concentration in mg/
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Answer:
0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters
Step-by-step explanation:
To solve this question, we need to understan 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:
Either the sample mean differs by 0.8 mm or less from the population mean, or it differs by more. The sum of these probabilities is decimal 1.
Probability it differs by less than 0.8mm
pvalue of Z when X = 139 + 0.8 = 139.8mm subtracted by the pvalue of Z when X = 139 - 0.8 = 138.2 mm
X = 139.8
By the Central Limit Theorem
has a pvalue of 0.7995
X = 138.2
has a pvalue of 0.2005
0.7995 - 0.2005 = 0.5990
What is the probability that the sample mean would differ from the population mean by more than 0.8 millimeters?
p + 0.5990 = 0.4010
0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters