Mathematics, 03.10.2019 00:30 ErrorNameTaken505

Correct answer only !

use technology or a z-score table to answer the question.

the lengths of green beans for sale at a supermarket are normally distributed with a mean of 11.2 centimeters and a standard deviation of 2.1 centimeters. consider a bag of 150 green beans.

how many green beans will be 13 centimeters or shorter?

Correct answer only ! use technology or a z-score table to answer the question.

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Answer from: jona187

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be centimeters.

follows a normal distribution with

mean $\mu = 11.2$ and standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that will be less than or equal to the z-score in question. (In case the question is asking for the probability that is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter

Enter the mean and standard deviation (or variance if required).Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives 0.804317 .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials

: 150.Probability of success

: 0.804317.

Let be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

answer: (yeees, you need a poet.)

//blooming serenity//

beautiful in the morning

mysterious at night

wonderous nature souring

with all it's mystical might

the flower that blooms

to the fox that looms

nocturnal creatures

with various features

a blooming flower soon in sight

as a new little hatchling fears to take flight

afraid to leave it's calming nest,

but dear little bird, mother knows best

as sure as there are stars in the sky,

life is not about who, where what, when or why-

it's the journey that follows

adventures we swallow

of nature's deep wonders,

natural disasters and storm's thunders.

blooming serenity

life and nature, truly is heavenly.

-kerstien lehman

step-by-step explanation:

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