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raketka [301]
3 years ago
15

A golfer's bag contains 24 golf balls, 18 of which are ProFlight brand and the other 6 are DistMax brand. Find the probability t

hat he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax.
Mathematics
1 answer:
eduard3 years ago
8 0

Answer:

1. Assuming with replacement, the probability is 7.91%; or

2. Assuming without replacement, the probability is 8.64%

Step-by-step explanation:

Total number of golf balls = 24

Let Pr denotes probability, P denotes ProFlights, D denotes DistMax.

The probability of selecting 5 balls can be with or without replacement. Since the question is silent on this, the answers to the methods are provided as follows:

1. Assuming with replacement

Pr(4 P and 1 D)  = (18/24) × (18/24) × (18/24) × (18/24) × (6/24)

                         = 0.75 × 0.75 × 0.75 × 0.75 × 0.25

                         = 0.0791  = 7.91%

2. Assuming without replacement

Here, we assume that 4 ProFlights are selected first before 1 DistMax is selected, and the probability is as follows:

Pr(4 P and 1 D) = (18/24) × (17/23) × (16/22) × (15/21) × (6/20)

                        = 0.7500 × 0.7391  × 0.7273  × 0.7143  × 0.3000  

                        =  0.0864  = 8.64%

Therefore, the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax is 7.91% assuming with replacement or 8.64% assuming without replacement.

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I hope this helps!

3 0
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Answer:

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7 0
3 years ago
Which function has a simplified base of 4? f(x) = 2 f(x) = 2 f(x) = 4 f(x) = 4
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4 years ago
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A data set lists earthquake depths. The summary statistics are
irinina [24]

Answer:

Step-by-step explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean \overline x = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean \mu = 6.00

Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

To start with the hypothesis;

The null and the alternative hypothesis can be computed as :

H_o: \mu = 6.00 \\ \\  H_1 : \mu \neq 6.00

The test statistics for this two tailed test can be computed as:

z= \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt {n}}}

z= \dfrac{6.86 - 6.00}{\dfrac{4.37}{\sqrt {400}}}

z= \dfrac{0.86}{\dfrac{4.37}{20}}

z = 3.936

degree of freedom = n - 1

degree of freedom = 400 - 1

degree of freedom = 399

At the level of significance ∝ = 0.01

P -value = 2 × (z < 3.936)  since it is a two tailed test

P -value = 2 × ( 1  - P(z ≤ 3.936)

P -value = 2 × ( 1  -0.9999)

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P -value =  0.0002

Since the P-value is less than level of significance , we reject H_o at level of significance 0.01

Conclusion: There is sufficient evidence to conclude that the original claim that the mean of the population of earthquake depths is  5.00 km.

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