Answer :
Using the normal distribution, it is found that the goal of between 1 and 1.2 ounces of ketchup on each burger will happen 70.55% of the time.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is given by the following rule:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure X is above or below the mean, depending if the z-score is positive or negative.
- From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X.
The mean and the standard deviation of the ketchup amounts in this problem are given as follows:
[tex]\mu = 1.05, \sigma = 0.08[/tex]
The proportion of time that the goal will be met is the p-value of Z when X = 1.2 subtracted by the p-value of Z when X = 1, hence:
X = 1.2:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (1.2 - 1.05)/0.08
Z = 1.88
Z = 1.88 has a p-value of 0.9699.
X = 1:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (1- 1.05)/0.08
Z = -0.63
Z = -0.63 has a p-value of 0.2644.
0.9699 - 0.2644 = 0.7055.
Hence the percentage is of 70.55%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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