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zaharov [31]
2 years ago
8

2. A statistics student plans to use a TI-84 Plus calculator on her final exam. From past experience, she estimates that there i

s 0.92 probability that the calculator will work on any given day. Because the final exam is so important, she plans to use redundancy by bringing in two TI-84 Plus calculators. What is the probability that she will be able to complete her exam with a working calculator? Does she really gain much by bringing in the backup calculator? Explain. [6 points]
Mathematics
1 answer:
Anarel [89]2 years ago
7 0

Answer:

  1. P(≥1 working) = 0.9936
  2. She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.

Step-by-step explanation:

1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...

... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936

2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.

If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.

This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.

_____

My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)

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Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug
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Answer:

4.979044478499338 × 10²⁶

Step-by-step explanation:

Combination can be used to determine the number of ways the mice can be selected for the drugs (A, B) and the control group.

Combination factorial is define by  ⁿCr = \frac{n!}{(n-r)!r!}

21 group of mice receiving Drug A  can be selected in ⁶⁰C₂₁ = \frac{60!}{21!39!}

(60 - 21 = 39 ) mice remained for selection of 21 mice for the second drug

Drug B 21 mice can be chosen with ³⁹C₂₁ = \frac{39!}{21!18!}

( 39 - 21 = 18)  remained for control with ¹⁸C₁₈ = \frac{18!}{18!0!}

The number of ways the mice can be chosen for drug A, drug B and the control = ⁶⁰C₂₁ × ³⁹C₂₁ × ¹⁸C₁₈ = \frac{60!}{21!39!} × \frac{39!}{21!18!} × \frac{18!}{18!0!} = 4.979044478499338 × 10²⁶

8 0
2 years ago
A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tes
jeka94

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis  

H_{0}: p = 0.05\\H_A: p > 0.05

This is a one-tailed(right) test.  

Formula:

\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08

z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Putting the values, we get,

z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

5 0
1 year ago
A study was conducted to determine the mean birth weight of a certain breed of kittens. Consider the birth weights of kittens to
kipiarov [429]
3.56 +/- 0.011 = ( 3.549, 3.571). This is the answer.
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2 years ago
The system shown has the unique solution (2, y, z). Solve the system and select the values that complete the solution. y = 0 y =
madreJ [45]
So we are given a system:
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8 0
2 years ago
Read 2 more answers
A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc
lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

8 0
2 years ago
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