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erica [24]
3 years ago
12

A hiker attempting to cross a wilderness area has two GPS systems of intermittent reliability because of the areas remoteness .f

rom past experience he estimates that at any moment the probability that system I will work is 0.85, that system II will work is 0.90 , and both will work is 0.80 . the probability that at any moment at least one of the systems will work is about___________.A. 0.87
B. 0.92
C. 0.75
D. 0.95
E. 0.88
Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

D. 0.95

Step-by-step explanation:

The probability that at any moment, at least one of the systems will work is determined by the probability that system I works, added to the probability that system II works, and subtracted by the probability that both systems work:

P(S_I\cup S_{II}) =P(S_I)+P(S_{II})-P(S_I\cap S_{II})\\P(S_I\cup S_{II}) =0.85+0.90-0.80\\P(S_I\cup S_{II}) =0.95

The probability that at any moment at least one of the systems will work is about 0.95.

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Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v
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Answer:

\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

\bar{x} = \$71  

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

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=STDEV(number1, number2,....)

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 s = \$ 22.35

The confidence interval is given by

\text {confidence interval} = \bar{x} \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\

So the required 95% confidence interval is

\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

7 0
3 years ago
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The heights of American men are normally distributed. If a random sample of American men is taken and the confidence interval is
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Answer:

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