The masses of the oranges on sale at a farm stand are normally distributed with a mean of 239 grams and a standard deviation of 25 grams. enter the z-score of an orange that has a mass of 259 grams. update: the answer is 0.8

In order to calculate the z-score of a normal distribution, you need to apply the formula:

z = (o - m) / s

where:
o = observed value
m = mean
s = standard deviation

In your case:

z = (259 - 239) / 25 = 0.8

The answer is, indeed, z = 0.8

(-14+3/2b) - (1+8/2b) = -5 -5/2b(-10+b) + (7b-5) = 8b-15(7/2b-3) - (8+6b) = -5/2b-11(5+2b) + (2b+3/2) = 4b+ 13/2