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yuradex [85]
3 years ago
11

Some electronic devices are better used than new: the failure rate is higher when they are new than when they are six months old

. For example, half of the personal music players of a particular brand have a flaw. Of the player has the flaw, it dies in the first two months with a probability of 30%, in the first four months with a probability of 70%, and in the first six months with a probability of 100%. If the player DOES NOT have the flaw, however, it dies in the first two months with a probability of 0%, in the first four months with a probability of 5%, and in the first six months with a probability of 10%.
The probability that it has the flaw is ____.
Mathematics
1 answer:
Reptile [31]3 years ago
7 0
Given that half of the personal music players sold by a particular brand have a flaw. If the player has the​ flaw, it dies in the first six months. If it does not have this​ flaw, then only 20​% fail in the first six months.
Let A be the event that it fails in I 6 months
B1 = it has a flaw
B2 = it does not have a flaw
B1 and B2 are mutually exclusive and exhaustive
the chances that it has this​ flaw
=The probability that it has the flaw is

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A data set lists earthquake depths. The summary statistics are nequals300​, x overbarequals5.89 ​km, sequals4.44 km. Use a 0.01
kirza4 [7]

Answer:

Null hypothesis H_0:\mu=5.00km

Alternative hypothesis H_1:\mu\neq 5.00km

The p value is 0.000517, which is less than the significance level 0.01, therefore we reject the null hypothesis and conclude that population mean is not equal to 5.00.

Step-by-step explanation:

It is given that a data set lists earthquake depths. The summary statistics are

n=300

\overline{x}=5.89km

s=4.44km

Level of significance = 0.01

We need to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 5.00.

Null hypothesis H_0:\mu=5.00km

Alternative hypothesis H_1:\mu\neq 5.00km

The formula for z-value is

z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}

z=\frac{5.89-5.00}{\frac{4.44}{\sqrt{300}}}

z=\frac{0.89}{0.25634351952}

z=3.4719

The p-value for z=3.4719 is 0.000517.

Since the p value is 0.000517, which is less than the significance level 0.01, therefore we reject the null hypothesis and conclude that population mean is not equal to 5.00.

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