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guapka [62]
3 years ago
9

Suppose that a certain brand of light bulb has a mean life of 450 hours and a standard deviation of 73 hours. Assuming the data

are bell-shaped: (Show work to get full credit)
a. Would it be unusual for a light bulb to have a life span of 320 hours? 615 hours? Justify each response.
b. According to the Empirical Rule, 99.7% of the light bulbs have a lifetime between what two values?
c. Determine the percentage of light bulbs that will have a life between 304 and 596 hours.
Mathematics
1 answer:
Mrac [35]3 years ago
4 0

Answer:

yes it is correct

Step-by-step explanation:

plz give brainliest.

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alexandr402 [8]
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</span><span>2. The scale of a map is 7 in = 16 mi map: 4.9 in actual: ______ mi
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</span><span>3. The scale factor for a model is 5 cm = ________ m Model : 72.5 cm actual: 165.3 m
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</span><span>5. The scale of a map is 1 ft = 9.6 mi map: ________ ft actual: 38.4 mi
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3 years ago
Roulette is a casino game that involves players betting on where a ball will land on a spinning wheel. An American roulette whee
nikitadnepr [17]

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elixir [45]
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A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by
omeli [17]

Answer:

1) A. H0: p = 0.30

HA: p not equal to 0.30

2) A. The Independence Assumption is met.

C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

P-value = 0.0367

4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical​ insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

H_0: \pi=0.3\\\\H_a:\pi\neq 0.3

2) Cheking the conditions:

The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089

This test is a two-tailed test, so the P-value for this test is calculated as:

\text{P-value}=2\cdot P(z>2.089)=0.0367

As the P-value (0.0184) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a significant change in the percent of bills being paid by medical​ insurance.

5 0
3 years ago
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