The distribution of actual weights of wedges of cheddar cheese produced at a dairy is normal with a mean of 11.9 ounces and a standard deviation of 0.2 ounces.

Question

(Round all answers to 4 decimal places, if needed.)

(a) The probability that a randomly chosen wedge of cheddar cheese is greater than 11.83 is .

(b) If a sample of 16 is randomly chosen, then the distribution of the sample mean weight is (approximately normal/not normal/left-skewed/right-skewed) with a mean of  and a standard deviation of .

(c) The probability that the sample mean weight of this sample of 16 is less than 11.83 is .

(d) The probability that the sample mean weight of this sample of 16 is greater than 11.83 is .

(e) The probability that the sample mean weight of this sample of 16 is between 11.83 and 11.98 is .

(f) There is only a 3% chance that the average weight of a sample of these 16 cheese wedges will be below .

Answer 1

μ = 11.9 oz

σ = 0.2 oz

Use the z-score formula to help you find the probability from the z-score table in your textbook.

z = (X - μ) / σ

a.) z = (11.83 - 11.9) / 0.2 = -0.35 ⇒ P(z > -0.35) = 1 - P(z ≤ -0.35) = 1 - 0.3632 = 0.6368

The probability that a randomly chosen wedge of cheddar cheese is greater than 11.83 oz is 0.6368.

b.) If a sample of 16 is randomly chosen, then the distribution of the sample mean weight is approximately normal with a mean of 11.9 and a standard deviation of σ/√n = 0.2/√16 ≈ 0.05.

c.) z = (11.83 - 11.9) / 0.05 = -1.4 ⇒ P(z < -1.4) = 0.0808

The probability that the sample mean weight of this sample of 16 is less than 11.83 is calculated similarly to part (a), but using the standard deviation of the sample mean (0.05). Therefore, it is about 0.0808.

d.) z = (11.83 - 11.9) / 0.05 = -1.4 ⇒ P(z > -1.4) = 1 - 0.0808 = 0.9192

The probability that the sample mean weight of this sample of 16 is greater than 11.83 is about 0.9192.

e.) z₁ = (11.83 - 11.9) / 0.05 = -1.4 ⇒ P(z₁ < -1.4) = 0.0808

z₂ = (11.98 - 11.9) / 0.05 = 1.6 ⇒ P(z₂ < 1.6) = 0.9452

P(-1.4 < z < 1.6) = 0.9452 - 0.0808 = 0.8644

The probability that the sample mean weight of this sample of 16 is between 11.83 and 11.98 is about 0.8644.

f.) There is only a 3% chance that the average weight of a sample of these 16 cheese wedges will be below a certain value. This value can be found by looking up the z-score corresponding to a 3% probability in a standard normal distribution table (approximately -1.88) and solving for X:

X = μ + z*σ = 11.9 - (1.88)*(0.05) ≈ 11.806

There is a 3% chance that the average weight of a sample of 16 cheese wedges will be below 11.806 ounces.