Question

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as the population parameters). Suppose 36 pro golfers played the course today. Find the probability that the average score of the 36 golfers was between 70 and 71. (Hint: think of this in terms of a sampling distribution with sample size 36) Group of answer choices

Answer 1

Answer:

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$

$P(70 \leq x \leq 71) = 47.7\%$

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

Answer 2

The probability that the average score of the 36 golfers was between 70 and 71 is 47.7%.

What is normal a distribution?

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as the population parameters).

Suppose 36 pro golfers played the course today.

The probability that the average score of the 36 golfers was between 70 and 71 will be

$z-score = \dfrac{x - \mu}{\sigma}$

Then

$Standard \ error \ of\ sampling = \dfrac{\sigma}{\sqrt{n}} = \dfrac{3}{\sqrt{36}} = 0.5$

$P(70\leq x\leq 71) = P(\dfrac{70-70}{\frac{3}{\sqrt{36}}}\leq z\leq \dfrac{71-70}{\frac{3}{\sqrt{36}}}) = P(0\leq z\leq 2)\\\\P(70\leq x\leq 71) = P(z\leq 2 ) - P(z\leq 0) = 0.977 - 0.500 = 0.477 = 47.7 \%$

More about the normal distribution link is given below.

https://brainly.com/question/12421652