I need help with a problem from my statistics homework. I don't understand this...

Question

A particular fruit's weights are normally distributed, with a mean of 640 grams and a standard deviation of 26 grams.

If you pick one fruit at random, what is the probability that it will weigh between 598 grams and 684 grams

Answer 1

Let x be fruits weight. You are trying to find probability x will weight between 598 and 684 grams:

P(598 < x < 684)

When mean is 640 and standard deviation is 26.

To find that probability we could use the normcdf function on the TI calculator:

normcdf(598,684,640,26)

Or we could convert the end points of that range to standardized z-scores, where z-score = ((x - mean)/ standard deviation):

To convert 598 to z-score: (598-640)/26 = -1.615

To convert 684 to z-score: (684-640)/26 = 1.692

So that range becomes:

P(-1.615 < z < 1.692)

The distribution of the standardized z-scores has mean 0 and standard deviation 1.

To find the above probability we could use the normcdf function on the TI calculator:

normcdf(-1.615,1.692,0,1)

Or you can use the standard normal probability table. Since probability table is cumulative up to z-score on the left we can compute that probability as:

P(z < 1.692) - P(z < -1.615) = 0.955 - 0.269 = 0.686