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IceJOKER [234]
3 years ago
14

In order to attract more advertisers to the arena during the sports events, the arena plans to offer advertising space as corpor

ate prizes. 720 companies enter. 8 companies win. What is the probability of a company winning? give your answer as a fraction
Mathematics
2 answers:
beks73 [17]3 years ago
3 0
720 divided by 8 and then look up the answer to that as a fraction on Google
Jobisdone [24]3 years ago
3 0

Answer: \dfrac{1}{90}

Step-by-step explanation:

Given : The total number of companies = 720

The number of companies won = 8

Now, the probability of a company winning is given by ;-

\text{P(winning)}=\dfrac{\text{Number of company wins}}{\text{ Total companies}}\\\\\Rightarrow\ \text{P(winning)}=\dfrac{8}{720}\\\\\Rightarrow\ \text{P(winning)}=\dfrac{1}{90}

Hence, the probability of a company winning  is \dfrac{1}{90}.

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A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p
Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

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