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Sagot :
Answer:
0.1587 = 15.87% probability that the weight of a randomly selected candy bar is less than 8 ounces.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces.
This means that [tex]\mu = 8.1, \sigma = 0.1[/tex].
(a) Find the probability that the weight of a randomly selected candy bar is less than 8 ounces
This is the pvalue of Z when X = 8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8 - 8.1}{0.1}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
0.1587 = 15.87% probability that the weight of a randomly selected candy bar is less than 8 ounces.
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