A and b are positive integers and a–b = 3. Evaluate the following:
1 answer:
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Answer:
a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b) The weight that 80% of the apples exceed is of 78.28g.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.
This means that
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
This is 1 subtracted by the p-value of Z when X = 100. So
has a p-value of 0.9697
1 - 0.9696 = 0.0304
0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?
This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.
The weight that 80% of the apples exceed is of 78.28g.