A population of values has a normal distribution with μ = 107.7 and σ = 16.8. You intend to draw a random sample of size n = 243. Find the probability that a single randomly selected value is greater than 107.2. P (X > 107.2) = 0.5119 Correct Find the probability that a sample of size n = 243 is randomly selected with a mean greater than 107.2. P (M > 107.2) = 0.5119 Incorrect Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

1. P (x>107,2) = 0,5119

2. P (x (bar) > 107,2) = 0,6772

Step-by-step explanation:

Hello!

dа ta:

The population of values with normal distribution

μ: 107,7

σ: 16,8

n=243

1. According to the text: X≈N (μ; σ2)

To standardize it and calculate the asked probability we can use is

Z = (x-μ) / σ≈N (0; 1)

First, let's rewrite the probability to its complement since most of the tables of probability accumulate from left to right

P (X>107,2) = 1 - P (X≤107,2)

Then we standardize

1-P (Z≤ (x-μ) / σ) = 1 - P (Z≤ ((107,2-107,7) / 16,8) = 1 - P (Z≤-0,029) = 1 - P (Z≤-0,03)

= 1 - 0,48803 = 0,51197

2. The next probability asked is not about a random value from the sample (x) but for a value that the sample mean might take. To calculate this probability we need to take the distribution of the sample mean into consideration.

This is x (bar) ≈ N (μ; δ2/n)

For standardization, we will also use the Z distribution, but under the distribution of the sample mean. Since the mean is for the same population, the values that μ and δ take are the same, but in this case, the sample n also plays a role in the formula.

In this case the statistic will be Z = (x (bar) - μ) / δ/√n ≈N (0; 1)

For the probability

P (x (bar) > 107,2) = 1 - P (x (bar) ≤107,2)

1 - P (Z ≤ (x (bar) - μ) / δ/√n) = 1 - P (Z ≤ (107,2 - 107,7) / 16,8/√243) = 1 - P (Z ≤ - 0,46)

1 - P (Z ≤ - 0,46) = 1 - 0,32276 = 0,67724

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