I literally dont under this pls help
1 answer:
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Answer:
The probability of choosing a blue or yellow highlighter when one yellow highlighter is picked is 17/25.
Step-by-step explanation:
Here, according to the question:
Total number of yellow highlighters = 12
Total number of blue highlighters = 6
Total number of green highlighters = 4
Total number of orange highlighters = 3
So, the total number of highlighters = 12 + 6 + 4 + 3 = 25
Let E1 : Event of selecting a yellow highlighter.
P(E1) = =
Let E2 : Event of selecting blue or yellow highlighter ONCE yellow highlighter is selected.
So, the total yellow highlighters left after selecting one = 12 -1 = 11
Also, the number of blue highlighter = 6
So, the TOTAL FAVORABLE options to E2 = 6+ 11 = 17
P(E2) =
=
Hence, the probability that a blue or yellow highlighter is selected given that a yellow highlighter is selected is 17/25.
Answer:
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Step-by-step explanation:
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.
At the null hypothesis, we test if the mean cost is still the same, that is:
At the alternative hypothesis, we test if the mean cost has increased, that is:
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question
In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
13252 is tested at the null hypothesis:
This means that
The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.