A particular fruit's weights are normally distributed, with a mean of 405 grams and a standard deviation of 19 grams. If a fruit is picked at random then 17% of the time, its weight will be greater than how many grams?

413.126

Step-by-step explanation:

Given a normal distribution :

Mean (m) = 405

Standard deviation (s) = 19

Weight of fruit will be greater than how many grams (x) if picked 17% of the time:

P(Z > x) = 0.17 ;

The Zscore for P(Z > x) = 0.17 equals 0.954 (Z probability calculator)

Zscore = (x - m) / s

0.954 = (x - 405) / 19

0.954 * 19 = x - 405

18.126 = x - 405

x = 18.126 + 405

x = 423.126

Weight will be greater than 423.126

on your paper it doesn't say the answer but the the answer is altitude

step-by-step explanation:

i dony know

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