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1 year ago

What is the probability that in a randomly selected game, the first goal is scored in less than 4 minutes?

1 answer:

4 0

Based on the characteristics of the distribution, the probability that the first goal is scored in less than 4 minutes is 0.

<h3>What is the probability?</h3>

For us to be able to calculate the probability that the first goal is scored in less than 4 minutes, the distribution needs to satisfy the requirements of the Central Limit Theorem(CLT).

Based on the fact that the distribution here is extremely right skewed, then it does not satisfy the requirements of CLT so the probability is 0.

First part of question:

The distribution of the time it takes for the first goal to be scored in a hockey game is known to be extremely right skewed with population mean 12 minutes and population standard deviation 8 minutes.

Find out more on the requirements of the Central Limit Theorem at brainly.com/question/18403552

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An equation in the system is y = x + 1. What is the other equation in the system if the solution is only at (0, 1)?

Setler [38]

x + y = 0 hope this helps

6 0

3 years ago

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0

3 years ago

On a number line, what is x-4<-3?

goblinko [34]

Add 4 to both sides

x - 4 < -3

x-4+4 < -3+4

x < 1

To graph this on a number line, plot an open circle at 1 on the number line. Do not fill in the open circle. Shade to the left of the open circle. The shaded region represents all x values smaller than 1.

The graph is shown below.