Thread: Normal distribution

A container of 20ft carry a weight that is normally distributed with a mean of 5000 t and a standard deviation of 400 t. What percentage of container weight between 5400 and 5600 t?

This is where I reached, how should I conclude on the percentage?
For χ = 5400
Z₅₄₀₀ = (5400-5000)/400
Z₅₄₀₀ = 10
For χ = 5600
Z₅₆₀₀ = (5600-5000)/40
Z₅₆₀₀ = 15

First you need to check your calculations.

Thank you. Corrected as;
For χ = 5400
Z5400 = (5400-5000)/400
Z5400 = 1
For χ = 5600
Z5600 = (5600-5000)/400
Z5600 = 1.5

Good.

P(1<Z<1.5) = P(Z<1.5) - P(Z<1).

Do you understand why and can you find these probabilities?

here is where I get confusion, distribution table I have I read for example from z column down to locate 1.5 and I get 0.4332 and for 1 0.3413. Help me how to read the values of 1.5 and 1

Originally Posted by benedec

here is where I get confusion, distribution table I have I read for example from z column down to locate 1.5 and I get 0.4332 and for 1 0.3413. Help me how to read the values of 1.5 and 1

Those will do but they are P(0<Z<1.5) and P(0<Z<1).

The answer you need is still the difference between these two values.

was worried probably i read the table incorrectly, the difference i get is this; P(1<Z<1.5) = P(Z<1.5) – P(Z<1) = 0.4332 – 0.3413 = 0.0919 = 9.19%