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Andre45 [30]
3 years ago
12

Internet company Gurgle is carrying out testing on the efficiency of its search engine. A sample of 31 searches have been carrie

d out and the time taken to display the results has been recorded for each search. The mean search time for the sample was calculated as 0.2258 seconds. The standard deviation of the search times for the sample was calculated as 0.0188 seconds. The population standard deviation of search times is unknown.
1. Select all the techniques that are commonly used to construct a confidence interval for the mean when the population standard deviation (σ) is unknown:

a. Approximate the population standard deviation (σ) with the sample standard deviation (s)
b. Replace the sample size (n) with n-1
c. Decrease the confidence level to compensate for the increased margin of error
d. Approximate the standard normal distribution with the Student's t distribution

2. Calculate the upper and lower bounds of the 95% confidence interval for the mean search time for the Gurgle search engine.
Mathematics
1 answer:
Mrac [35]3 years ago
3 0

Answer:

d. Approximate the standard normal distribution with the Student's t distribution

(0.2199 ; 0.2327)

Step-by-step explanation:

Given that :

Sample size, n = 31

Sample mean, xbar = 0.2258

Sample standard deviation, s = 0.0188

Confidence interval (C. I) :

xbar ± margin of error

Margin of Error : Tcritical * s/sqrt(n)

Degree of freedom, df = n - 1 = 31 - 1 = 30

Tcritical value :

T0.05/2, 30 = 2.042

Margin of Error = 2.042 * 0.0188/sqrt(31)

Margin of Error = 0.0068949

C. I = 0.2258 ± 0.0068949

Lower boundary : (0.2258 - 0.006895) = 0.2189

Upper boundary : (0.2258 - 0.006895) = 0.2327

(0.2199 ; 0.2327)

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Make a proportion

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The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
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The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

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b. A componenet fails in the interval from 1000 to 2000 hours.

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Answer: a. P(x>3000) = 0.5

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              c. P(x<1000) = 0.6321

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Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

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P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

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