Question

He weight of cans of fruit is normally distributed with a mean of 1,000 grams and a standard deviation of 50 grams. what percent of the canss weigh 860 grams or less?

Answer 1

To calculate these values, you need first the probabilities P(X < 860), P(X < 1055) and P(X < 1100) from the cumulative normal probability distribution for the variable X (the weight of a can of pears, in this case). It's too tedious to work this out from the equation for this distribution and the given values for its mean μ and standard deviation σ; the simple method is to reduce the required values of X to the reduced parameter z using the relationship 

z = (X - μ)/σ 

which in this case gives X = 860 → z = -2.80, X = 1055 → z = +1.10 and X = 1100 → z = +2.00. 

You then look up the fractional probabilities for the required z values in a published table for the cumulative standard normal distribution (links to two are provided below). To save space, these give one set of values for both positive and negative values of z; to convert the tabulated numbers P(z) to the correct probabilities 

- for positive z, add 0.5 to the tabulated value P(z) 
- for negative z, subtract the corresponding P(z) value for positive z from 0.5 

for z = 2.80, P(z) = 0.4974, so P(z < -2.80) = 0.5 - 0.4974 = 0.0026 
for z = 1.10, P(z) = 0.3643 and P(z < 1.10) = 0.3643 + 0.5 = 0.8643 
for z = 2.00, P(z) = 0.4772 and P(z < 2.00) = 0.4772 + 0.5 = 0.9772. 

Finally we can get the results we require 

a. P(X < 860) = P(z < -2.80) = 0.0026 = 0.26% 

b. P(1055 < x < 1100) = P(z < 2.00) - P(z < 1.10) = 0.9772 - 0.8643 = 0.1129 = 11.29% 

c. P(860 < X < 1055) = P(z < 1.10) - P(z < -2.80) = 0.8643 - 0.0026 = 0.8617 = 86.17%