Answer:
B) 8,4 pounds
Step-by-step explanation:
You substitute <em>four</em><em> </em><em>weeks</em><em> </em>in for <em>x</em>,<em> </em>multiply that by 1,8 to get 7,2, and finally add 1,2 to it, to get the total weight loss of 8,4 pounds [3,81018 kilograms].
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Answer:
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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.
A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.
This means that
a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.
pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 - 0.05 = 199.9.
X = 200.9
has a pvalue of 0.8413
X = 199.9
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.