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The weight of an energy bar is approximately normally distributed with a mean of 42.65 grams and a standard deviation of 0.045 grams.

(a) What is the probability that an individual energy bar weighs less than 42.615 grams? (Round to three decimal places as needed.)

(b) If a sample of 4 energy bars is selected, what is the probability that the sample mean weight is less than 42.615 grams? (Round to three decimal places as needed.)

(c) If a sample of 25 energy bars is selected, what is the probability that the sample mean weight is less than 42.615 grams? (Round to three decimal places as needed.)

(d) Explain the difference in the results of (a) and (c).

(e) Explain the difference in the results of (b) and (c).

1 Answer

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Final answer:

In summary, the probability of an individual energy bar weighing less than 42.615 grams is approximately 0.218. The probability that a sample mean weight of 4 energy bars is less than 42.615 grams is approximately 0.236. The probability that a sample mean weight of 25 energy bars is less than 42.615 grams is approximately 0.345.

Step-by-step explanation:

(a) To find the probability that an individual energy bar weighs less than 42.615 grams, we first need to calculate the z-score. The z-score formula is given by: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Plugging in the values, we get: z = (42.615 - 42.65) / 0.045 = -0.7778. Now, we can use a z-table or statistical software to find the corresponding probability, which is approximately 0.218.

(b) To find the probability that the sample mean weight of 4 energy bars is less than 42.615 grams, we need to calculate the standard error of the mean first. The standard error formula is given by: σ / sqrt(n), where n is the sample size. Plugging in the values, we get: 0.045 / sqrt(4) = 0.0225. Now, we can use the z-score formula (as in part a) to find the corresponding probability, which is approximately 0.236.

(c) To find the probability that the sample mean weight of 25 energy bars is less than 42.615 grams, we can follow the same steps as in part b, but with a different sample size. The standard error formula becomes: 0.045 / sqrt(25) = 0.009. Using the z-score formula, we find the probability to be approximately 0.345.

(d) The difference in the results of (a) and (c) can be explained by the fact that as the sample size increases, the standard error decreases, resulting in a narrower distribution and a smaller probability of obtaining extreme values.

(e) The difference in the results of (b) and (c) can be explained by the same reasoning as in part (d). As the sample size increases, the standard error decreases, leading to a narrower distribution and a smaller probability of obtaining extreme values.

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