I totally forgot how to find a z-score. Please help and explain.

Weights of the vegetables in a field are normally distributed. From a sample, Carl Cornfield determines the mean weight of a box of vegetables to be 180 oz. with a standard deviation of 8 oz. He wonders what percent of the vegetable boxes he has grouped for sale will have a weight more than 196 oz.

Carl decides to answer the following questions about the population of vegetables from these sample statistics:

Carl calculates the z-score corresponding to the weight 196 oz. (to the nearest tenth) ___
Using the table below, Carl sees the area associated with this z-score is 0.
Carl rounds this value to the nearest thousandth or 0.
Now, Carl subtracts 0.50 - 0. = %.

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sqdancefan sqdancefan · Answer 1 · 2018-09-03T23:02:41+02:00

The z-score is the number of standard deviations from the mean. For a given x-value, mean (μ), and standard deviation (σ) the z-score is given by

... z = (x -μ)/σ

(a) x=196, μ = 180, σ = 8 gives a z-score of

... z = (196 -180)/8 = 16/8

... z = 2.0

(b) The area associated with x=2 is 0.4772

(c) Rounded to the nearest thousandth, this is 0.477

(d) Carl's subtraction gives

... 0.50 - 0.477 = 2.3%

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mathmate mathmate · Answer 2 · 2018-09-03T23:38:35+02:00

Just a simple review:

Z(X)= (X-mean)/standard deviation=(X- μ ) / σ

Given:

μ = 180

&sigmal = 8

Z(196)=(196-180)/8 = 2

Here, we're looking for the right tail, i.e. for Z scores greater than 2.

The table gives the area (probability) under the normal curve from Z(0) to Z(X).

Thus for estmating probability for the right tail, we subtract the Z-score from 0.5.

The value of probability where

P(0<Z<2)=0.4772

P(Z>2.0) = 0.5-0.4772 = 0.0228

Therefore the probability of box weghing more than 196 oz is 0.0228, 0r 2.28%.