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The fill amount in 2-liter soft drink bottles is normally distributed with a mean of 2.0 liters and a mean of 0.05 litters. If bottles contain less than 95% of the listed net content (i.e., less than 1.90 litters) the manufacturer may be subject to penalty by the state office of consumer affairs. Bottles that have a net content above 2.10 liters may cause excessive spillage upon opening. What is the probability that a randomly selected bottle will contain:
a. Between 1.90 and 2.00 litters?
b. Between 1.90 and 2.10 litters?
c. Below 1.90 liters or above 2.10 litters?
About least how much soft drink (in litters) is contained in 99% of the bottles?
e. 99% of the bottles contain an amount that is between which two values, symmetric about the average?

Respuesta :

Answer:

a) 0.4722

b) 0.9545

c) 0.0455

d) 1.8837

e) between 1.8712 and 2.1288

Step-by-step explanation:

a) Z = (x – mean)/standard deviation

Le x be amount in liters in a bottle

P (1.9<x<2) = P((1.9– 2)/0.05 < z < (2– 2)/0.05)

P (1.9<x<2) = P( -2 < z < 0)

Then we use the z table to find the area under the curve.

P (1.9<x<2) = 0.4772

b) P (1.9<x<2.1) = P((1.9– 2)/0.05 < z < (2.1– 2)/0.05)

P (1.9<x<2.1) = P( -2 < z < 2)

Then we use the z table to find the area under the curve.

P (1.9<x<2.1) = 0.9545

c) There is a probability of 0.9545 that the content of a bottle is between 1.90 and 2.10 litters. Then the probability that the content is below 1.90 liters or above 2.10 litters is 1 – 0.9545 = 0.0455

d) z = (x– mean)/ standard deviation

x = z * standard deviation + mean

Now, z value for above the 99% has to be found using a z table.

In this case z = -2.326

x = -2.326 * .05 + 2

x = 1.8837

e) 99% of the bottles contain an amount that is between which two values, symmetric about the average means a 49.5% on both sides. Z values area -2.576 and 2.576.

Replacing in the formula

x = z * standard deviation + mean

x = -2.576 * .05 + 2

x = 1.8712

x = z * standard deviation + mean

x = 2.576 * .05 + 2

x = 2.1288

99% of the bottles contain an amount that is between 1.8712 and 2.1288

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