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2 years ago

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A sample of ultimate tensile strength observations (ksi) was taken. Use the accompanying descriptive statistics output from Mini

tab to calculate a 99% lower confidence bound for true average ultimate tensile strength, and interpret the result. (Round your answer to two decimal places.)

1 answer:

cricket20 [7]2 years ago

3 0

Answer:

The 99% lower confidence bound for the true average ultimate tensile strength is $\mu\geq134.53$ .

With these lower confidence bound we can claim that there is a 99% chance that the real ultimate tensile strength is greater than 134.53.

Step-by-step explanation:

The question is incomplete. The desctiptive statistics are the ones attached.

We have to calculate a 99% lower confidence bound for true average ultimate tensile strength.

For a 99% lower confidence bound, the z-value is z=2.33.

The margin of error is

$E=z\sigma/\sqrt{n}=2.33*4.59/\sqrt{153}=10.69/12.37=0.86$

Then, the 99% lower confidence bound for the true average ultimate tensile strength is:

$\mu\geq M-z\sigma/\sqrt{n}\\\\\mu\geq 135.39-0.86\\\\\mu\geq 134.53$

The experiment is probably done to estimate with a certain degree of confidence the minimum tensile strength of a population of elements, by evaluating a sample.

With these lower confidence bound we can claim that there is a 99% chance that the real ultimate tensile strength is greater than 134.53.

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2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

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Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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So to find the avrage u r going to add all the numbers then divide by the nuber of nubers u have there that will give u ur answer

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Dori and Malory are tracking their steps taken as a health goal. Dori leaves her house at 12:00 p.m. and walks at 50 steps per m

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Let n = minutes since 12:00 pm when Malory catches up to Dori.

Dori travels
(50 steps/min)*(n minutes) = 50n steps

Malory begins walking at 12:20 pm, so she walks for (n - 20) minutes. She travels
(90 steps/min)*(n - 20 min) = 90n - 1800 steps

Equate the steps traveled by Dori and Malory.
90n - 1800 = 50n
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The time corresponding to n = 45 min is 12:45 pm

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Answer:

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The statement tells us that first there is an expense of 0.4 cups of laundry detergent while washing one load of laundry and 0.7 coupon another load, therefore this means that the change in the bottle would be:

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In other words, the main bottle would have 1.1 cups less, that is, -1.1 cups.

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2 years ago

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