A particular fruit's weights are normally distributed, with a mean of 551 grams and a standard deviation of 20 grams. If you pick one fruit at random, what is the probability that it will weigh between 523 grams and 534 grams

11.69% probability that it will weigh between 523 grams and 534 grams

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

If you pick one fruit at random, what is the probability that it will weigh between 523 grams and 534 grams

This is the pvalue of Z when X = 534 subtracted by the pvalue of Z when X = 523. So

X = 534

has a pvalue of 0.1977

X = 523

has a pvalue of 0.0808

0.1977 - 0.0808 = 0.1169

11.69% probability that it will weigh between 523 grams and 534 grams

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"