Please help, i’m really struggling. will give brainliest

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

bija089 [108]

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.

8 0

4 years ago

....................

mylen [45]

Answer:

sub to juice_fox. ?....

3 0

2 years ago

Which expression illustrates the associative property of addition? (3 19) – 12 = (3 12) – 19 3 (19 – 12) = 3 (19 12) (3 19) – 12

Gre4nikov [31]

Answer:

Associative Property

For multiplication, the rule is "a(bc) = (ab)c"; in numbers, this means 2(3×4) = (2×3)4.

5 0

2 years ago

Danny poured 78 l of water into containers xy and z. the ratio of the volume of water in container x to the volume of water in c

Paladinen [302]

The ratio of x:y:z = 20:8:11

Total unit = 20+8+11 = 39

Value of 1 unit ratio = 78/39 = 2

So, the water in container X, 20(2) = 40 liters

Your answer is 40 liters

7 0

3 years ago

The results of a common standardized test used in psychology research is designed so that the population mean is 155 and the sta

galina1969 [7]

Answer:

The value 155 is zero standard deviations from the [population] mean, because $\\ x = \mu$ , and therefore $\\ z = 0$ .

Step-by-step explanation:

The key concept we need to manage here is the z-scores (or standardized values), and we can obtain a z-score using the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

z is the z-score.
x is the raw score: an observation from the normally distributed data that we want standardize using [1].
$\\ \mu$ is the population mean.
$\\ \sigma$ is the population standard deviation.

Carefully looking at [1], we can interpret it as the distance from the mean of a raw value in standard deviations units. When the z-score is negative indicates that the raw score, x, is below the population mean, $\\ \mu$ . Conversely, a positive z-score is telling us that x is above the population mean. A z-score is also fundamental when determining probabilities using the standard normal distribution.

For example, think about a z-score = 1. In this case, the raw score is, after being standardized using [1], one standard deviation above from the population mean. A z-score = -1 is also one standard deviation from the mean but below it.

These standardized values have always the same probability in the standard normal distribution, and this is the advantage of using it for calculating probabilities for normally distributed data.

A subject earns a score of 155. How many standard deviations from the mean is the value 155?

From the question, we know that:

x = 155.
$\\ \mu = 155$ .
$\\ \sigma = 50$ .

Having into account all the previous information, we can say that the raw score, x = 155, is zero standard deviations units from the mean. The subject earned a score that equals the population mean. Then, using [1]:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{155 - 155}{50}$

$\\ z = \frac{0}{50}$

$\\ z = 0$

As we say before, the z-score "tells us" the distance from the population mean, and in this case this value equals zero:

$\\ x = \mu$

Therefore

$\\ z = 0$

So, the value 155 is zero standard deviations from the [population] mean.

5 0

4 years ago