SOLUTION: An orange juice producer buys oranges from a large orange grove that has one variety of oranges. The amount of juice squeezed from these oranges is approximately normally​ distri

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Question 1203058: An orange juice producer buys oranges from a large orange grove that has one variety of oranges. The amount of juice squeezed from these oranges is approximately normally​ distributed, with a mean of 4.10 ounces and a standard deviation of 0.20 ounces. Suppose that you select a sample of 16 oranges.
a. What is the probability that the sample mean amount of juice will be at least 4.01 ounces?


b. The probability is 78​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? There is a 78​% probability that the sample mean amount of juice will be contained between ______​ounce(s) and _____ ​ounce(s) ​(Round to two decimal places as needed. Use ascending​ order.)


c. The probability is 73​% that the sample mean amount of juice will be greater than what​ value?

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
mean = 4.1 ounces.
standard deviation = .2 ounces.
sample size is 16 oranges.
standard error = standard deviation / square root of sample size = .2 / sqrt(16) = .05.

use the z-score formula of z = (x-m)/s
z is the z-score.
x is the sample mjean.
m is the population mean.
s is the standard error.

a. What is the probability that the sample mean amount of juice will be at least 4.01 ounces?

when x = 4.01, the formula becomes z = (4.01 - 4.1) / .05 = -1.8.
area to the right of that z-score is equal to .9641.

that's the probability that the mean of the sample will be at least 4.01 ounces.

b. The probability is 78​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? There is a 78​% probability that the sample mean amount of juice will be contained between ______​ounce(s) and _____ ​ounce(s) ​(Round to two decimal places as needed. Use ascending​ order.)

if the area between the two z-scores is .78, then the area outside of that range will be 1 - .78 = .22 / 2 = .11.

you will need 2 z-scores.
the high z-score will have an area to the left of it of 1 - .11 = .89
the low z-score will have an area to the left of it equal to .11.
using a z-score calculator, i get:
z-score with area of .89 to the left of it = 1.2265.
z-score with area of .11 to the left of it = -1.2265.
area in between will be .89 - .11 = .78.
use the z-score formula to find the raw score.
z = (x-m)/s
z is the z-score
x is the raw score = the mean of the smple
m is the mean
s is the standard error.
solve for x in this formula to get x = s * z + m.
when z = -1.2265 and m = 4.1 and s = .05, the formula becomes x = .05 * -1.2265 + 4.1 = 4.038675.
when z = .2265 and m = 4.1 and s = .05, the formula becomes x = .05 * 1.2265 + 4.1 = 4.161325.
the area between those two sample means will be equal to.78.

c. The probability is 73​% that the sample mean amount of juice will be greater than what​ value?

z-score will have an area of 1 - .73 = .27 to the left of it.
that z-score will be equal to -.6128.
the sample mean will be equal to x = z * s + m which would be equal to x = -.6128 * .05 + 4.1 = 4.06936.

the calculator at https://davidmlane.com/hyperstat/z_table.html can be used to confirm these results.

here are the results.

for part a.



for part b.



for part c.






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