Itotally forgot how to find a z-score. 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 carl rounds this value to the nearest thousandth or now, carl subtracts 0.50 - = %.

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%

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%.

i am sure it will be 90 degrees.

step-by-step explanation: