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Using the normal distribution, it is found that 10.56% of the apples weigh less than Jermaine's apple.
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In a <em>normal distribution</em> with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure X 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, which is the percentile of X, that is, the proportion of measures that are less than X.
In this problem:
- Mean of 0.33 pounds, thus
.
- Standard deviation of 0.06 pounds, thus
.
- The proportion that is less than 0.25 pounds is the p-value of Z when X = 0.25, thus:
has a p-value of 0.1056.
0.1056 x 100% = 10.56%.
10.56% of the apples weigh less than Jermaine's apple.
A similar problem is given at brainly.com/question/13411796