An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 4.90 ounces and a standard deviation of 0.20 ounce. Suppose that you select a sample of 25 oranges. a. What is the probability that the sample mean amount of juice will be at least 4.85 ounces? b. The probability is 70 % that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean? c. The probability is 71 % that the sample mean amount of juice will be greater than what value? a. The probability is nothing.

Question

Mathematics

Braden Owen

22 September, 02:50

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 4.90 ounces and a standard deviation of 0.20 ounce. Suppose that you select a sample of 25 oranges. a. What is the probability that the sample mean amount of juice will be at least 4.85 ounces? b. The probability is 70 % that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean? c. The probability is 71 % that the sample mean amount of juice will be greater than what value? a. The probability is nothing.

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Answers (1)

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Patrick · Answer 1

Step-by-step explanation:

Since the amount of juice squeezed from these oranges is approximately normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ) / σ/√n

Where

x = amount of juice squeezed.

µ = mean amount

σ = standard deviation

n = number of samples

From the information given,

µ = 4.90 ounces

σ = 0.20 ounce

n = 25

a) the probability that the sample mean amount of juice will be at least 4.85 ounces is expressed as

P (x ≥ 4.85) = 1 - P (x < 4.85)

For x = 4.85,

z = (4.85 - 4.9) / (0.2/√25) =

- 0.05/0.04 = - 1.25

Looking at the normal distribution table, the probability corresponding to the z score is 0.11

P (x ≥ 4.85) = 1 - 0.11 = 0.89

b) 70% = 70/100 = 0.7

Looking at the table, the two z scores corresponding to the probability value of 0.7 are 0.53 and - 0.53

For z = - 0.53,

- 0.53 = (x - 4.9) / (0.2/√25)

- 0.53 = (x - 4.9) / 0.04

x - 4.9 = 0.04 * - 0.53 = - 0.0212

x = - 0.0212 + 4.9 = 4.88

For z = 0.53,

0.53 = (x - 4.9) / (0.2/√25)

0.53 = (x - 4.9) / 0.04

x - 4.9 = 0.04 * 0.53 = 0.0212

x = 0.0212 + 4.9 = 4.92

c) P (x < 4.9)

The probability value is 71/100 = 0.71. Since the mean is greater, the probability value would be 1 - 0.71 = 0.29. Looking at the table, the z score corresponding to the probability value of 0.29 is - 0.55

Therefore,

- 0.55 = (x - 4.9) / (0.2/√25)

0.55 = (x - 4.9) / 0.04

x - 4.9 = 0.04 * 0.55 = 0.022

x = 0.022 + 4.9 = 4.922