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Sagot :
Answer:
15.87 %
Step-by-step explanation:
z = (.9-1)/.1 = -1
z score of -1 = .1587
The probability that a randomly selected small size package has a weight of less than 0.9 pounds is 15.87 %
We have given that,
A store sells ground meat in small, medium, and large sizes. The weights of the small size packages have a mean weight of 1 pound and a standard deviation of 0.1 pounds. If the distribution of weights for the small-size packages of ground meat is approximately normally distributed
What is the formula for the z-score?
[tex]Z = \frac{x - \mu}{\sigma}[/tex]
Z = standard score
x = observed value
[tex]\mu[/tex] = mean of the sample
[tex]\sigma[/tex] = standard deviation of the sample
z = (0.9-1)/0.1 = -1
z of -1 = 0.1587
Therefore the value of the z-score is 0.1587.
To learn more about the Z-score visit:
https://brainly.com/question/25638875
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