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

I hope you have a SUPER day!

step-by-step explanation:

one of the most important purposes of surface treatment is to alter surface properties according to the requirement of specific applications. for example, hydrophobic coating has been utilized on windshield and eyeglasses. gold atom monolayer on substrate has become a popular surface bioconjugation strategy.  

  answer:

  write 548 as 500 + 40 + 8. multiply each addend by 5. then add the three products to get 2,740.