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zavuch27 [327]
3 years ago
9

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated b

y a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)
________ seconds or less
Mathematics
1 answer:
Gala2k [10]3 years ago
3 0

Answer:

214.5 seconds or less time will qualify individuals for such training

Step-by-step explanation:

given that the time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec.

i.e. X the time that it takes a randomly selected job applicant to perform a certain task is N(150,25)

The fastest 10% are the ones who are above the 90th percentile

We know 90th percentile for Z std normal score is 1.28

Corresponding X score would be

x=\mu+1.28 \sigma\\X= 150+1.28(25)\\= 214.50

Round off to one decimal

214.5 seconds or less time will qualify individuals for such training

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We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

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where, \bar X = sample mean lifespan = 46,500 miles

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            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

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The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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