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Liono4ka [1.6K]
3 years ago
7

The response time for ski patrol rescue responders is measured by the length of time from when the radio call is finished and wh

en the responders locate the skier. Responders consider between 0 to 5 minutes as an ideal response time.
Supposing gathered data showed a Normal distribution with a mean of 6 minutes and standard deviation of 1.2 minutes, what percent of responses is considered ideal? Round to the nearest whole percent.
Mathematics
1 answer:
svp [43]3 years ago
4 0

Answer:

20

Step-by-step explanation:

just did ssignment

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Question Help
11111nata11111 [884]

Answer:

The null hypothesis is H_0: p \geq 0.2

The alternative hypothesis is H_1: p < 0.2

The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.

Step-by-step explanation:

Test the claim that the return rate is less than 20%.

At the null hypothesis, we test if the return rate is of at least 20%, that is:

H_0: p \geq 0.2

At the alternative hypothesis, we test if the return rate is less than 20%, that is:

H_1: p < 0.2

Normal distribution as an approximation to the binomial distribution to find the Z-score.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

0.2 is tested at the null hypothesis. Sample of 6957.

This means that p = 0.2, n = 6957. So

\mu = E(X) = np = 6957*0.2 = 1391.4

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6957*0.2*0.8} = 33.36

1319 surveys returned

This means that, due to continuity correction, X = 1319 + 0.5 = 1319.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1319.5 - 1391.4}{33.36}

Z = -2.16

P-value of the test:

The p-value of the test is the probability of 1319 or less people returning the survey, which is the p-value of Z = -2.16.

Looking at the z-table, Z = -2.16 has a p-value of 0.0154.

The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.

7 0
3 years ago
-7/8 - ( -3/16 ) write your answer in simplest from
ArbitrLikvidat [17]

Answer:

-7/8 - ( -3/16 )

-7/8 + 3/16

8 0
3 years ago
Read 2 more answers
PLEASE HELP!
butalik [34]
We know for the problem that the performer earned $120 at a performance where 8 people attend. We also know that he u<span>ses 43% of the money earned to pay the costs involved in putting on each performance, so we need to find the 43% of $120. To do that, we are going to divide 43% by 100%, and then multiply it by $120:
</span>\frac{43}{100}*120=51.6
Now we know that the performer uses $51.6 of $120 to pay the costs involved in putting on each performance. The only thing left to find his profits is subtract $51.6 from $120:
120-51.6=68.4

We can conclude that the performer makes a profit of $68.4 when 8 people attend his performance.
 
3 0
4 years ago
I'm GIVING POINTS OUT TO THOSE WHO DO THESE CHARACTERS: 0000000000000000000
hoa [83]

Thank you! Can I also have brainliest plsssss!!!!

8 0
3 years ago
Sunday morning, the temperature in Bayfield, Wisconsin was -15.6 degrees Fahrenheit. If the temperature rose 3.8 degrees Fahrenh
Ostrovityanka [42]

Answer:

just add them and leave it as a positive

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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