Apopulation 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!

Data:

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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i think the answer is true

step-by-step explanation: