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balandron
2 months ago
7

A PGA (Professional Golf Association) tournament organizer is attempting to determine whether hole (pin) placement has a signifi

cant impact on the average number of strokes for the 13th hole on a given golf course. Historically, the pin has been placed in the front right corner of the green, and the historical mean number of strokes for the hole has been 4.25, with a standard deviation of 1.6 strokes. On a particular day during the most recent golf tournament, the organizer placed the hole (pin) in the back left corner of the green. 64 golfers played the hole with the new placement on that day. Determine the probability of the sample average number of strokes exceeding 4.75
Mathematics
1 answer:
Leona [12.6K]2 months ago
8 0
Utilizing the complement rule alongside the standard normal table or Excel, we arrive at the results. To summarize previous concepts, the normal distribution represents a symmetrical probability distribution centered around the mean, highlighting that values near the mean occur more frequently than those further away. The Z-score acts as a statistical measure that indicates a value's relation to the average of a dataset, expressed in standard deviations from the mean. In addressing the problem: Let X denote the random variable representing population heights, and we regard the distribution for X as stipulated. For a sample size of n = 64, since the distribution of X holds normality, so does the distribution of the sample mean. We aim to determine the following probability, utilizing the Z-score derived via the specified formula.
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Yesenia swims laps at a pool. For every 10 laps she swims the backstroke, she swims 15 laps of freestyle.
Svet_ta [12734]

Response:

A 2:3

Detailed Explanation:

10:15

which is equivalent to 10/15

which simplifies to 2/3

I hope this assists you

<3

Red

4 0
2 months ago
Read 2 more answers
Kevin plants a total of 72 flowers in equal rows he plants 6 rows of yellow flowers and 2 rows of red flowers how many flowers a
PIT_PIT [12445]
For the red flowers, there are 18 in each row. For the yellow flowers, there are 6 in each row.
5 0
2 months ago
How many solutions does the equation sin(5x) = 1/2 have on the interval (0, 2PI]
lawyer [12517]

Answer:

Step-by-step explanation:

Considering the equation

Sin(5x) = ½

5x = arcSin(½)

5x = 30°

Then,

The general formula for sin is

5θ = n180 + (-1)ⁿθ

Dividing throughout by 5

θ = n•36 + (-1)ⁿ30/5

θ = 36n + (-1)ⁿ6

The solution range is

0<θ<2π which means 0<θ<360

First solution

When n = 0

θ = 36n + (-1)ⁿθ

θ = 0×36 + (-1)^0×6

θ = 6°

When n = 1

θ = 36n + (-1)ⁿ6

θ = 36-6

θ = 30°

When n = 2

θ = 36n + (-1)ⁿ6

θ = 36×2 + 6

θ = 78°

When n =3

θ = 36n + (-1)ⁿ6

θ = 36×3 - 6

θ = 102°

When n=4

θ = 36n + (-1)ⁿ6

θ = 36×4 + 6

θ = 150

When n=5

θ = 36n + (-1)ⁿ6

θ = 36×5 - 6

θ = 174°

When n = 6

θ = 36n+ (-1)ⁿ6

θ = 36×6 + 6

θ = 222°

When n = 7

θ = 36n + (-1)ⁿ6

θ = 36×7 - 6

θ = 246°

When n =8

θ = 36n + (-1)ⁿ6

θ = 36×8 + 6

θ = 294°

When n =9

θ = 36n + (-1)ⁿ6

θ = 36×9 - 6

θ = 318°

When n =10

θ = 36n + (-1)ⁿ6

θ = 36×10 + 6

θ = 366°

When n = 10 surpasses the θ range

Thus, the solutions range from n =0 to n=9

Therefore, there are 10 solutions within the interval 0<θ<2π

4 0
2 months ago
The system of equations can be solved using linear combination to eliminate one of the variables.2x − y = −4→10x − 5y = −203x +
babunello [11817]
2x − y = −4 → 10x − 5y = −20
3x + 5y = 59 → 3x + 5y = 59
13x = 39..... this is derived from adding the two equations on the right.

13x = 39 can be used in place of 3x + 5y = 59 in the initial system and will yield the same result.
6 0
2 months ago
A bag contains chips of which 27.5 percent are blue. A random sample of 5 chips will be selected one at a time and with replacem
lawyer [12517]

Answer:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997

Step-by-step explanation:

It is known that the mean and standard deviation of the sampling distribution of the sample proportion(\hat{p}) are represented as follows:-

\mu _{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}

, where p= Population proportion and n = sample size.

Let p denote the proportion of blue chips.

According to the information provided, we have

p= 0.275

n= 5

Thus, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{ 0.275(1- 0.275)}{5}}\\\\=0.19968725547\approx0.1997

Therefore, you will have the mean and standard deviation for the sample proportion of blue chips for samples of size 5:

\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997

6 0
1 month ago
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