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Yanka [14]
3 years ago
14

Help me with this probability question for (7th grade math)​

Mathematics
2 answers:
Agata [3.3K]3 years ago
6 0

Answer:

15 times

Step-by-step explanation:

Probability is calculated as

probability = \frac{favourableoutcomes}{count}

Here a favourable outcome is an even number : 2, 4, 6 and the count is 6

P(even number) = \frac{3}{6} = \frac{1}{2}

expected even number = \frac{1}{2} × 30 = 15 times

ElenaW [278]3 years ago
4 0

There is a 3/6 possibility for an even number to get chosen.

Might be wrong

bye

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In a recent Dunkin Donuts (DD) commercial, it states that 58% of Americans prefer to drink their coffee. Of the five hundred ran
Sergio [31]

Answer:

We conclude that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised.

Step-by-step explanation:

We are given that in a recent Dunkin Donuts (DD) commercial, it states that 58% of Americans prefer to drink their coffee.

Of the five hundred randomly selected people who were asked if they preferred DD coffees, 325 said they did.

<em>Let p = proportion of Americans who prefer to drink DD coffee</em>

SO, <u>Null Hypothesis</u>, H_0 : p \leq 58%   {means that the proportion of Americans who prefer to drink DD coffee is actually less than or equal to what was advertised}

<u>Alternate Hypothesis</u>, H_A : p > 58%   {means that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

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

where,  \hat p = sample proportion of people who prefer to drink DD coffee in a sample of 500 people = \frac{325}{500} = 65% or 0.65

            n = sample of people = 500

So, <em><u>test statistics</u></em>  =   \frac{0.65-0.58}{{\sqrt{\frac{0.65(1-0.65)}{500} } } } }

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<em>Now at 0.05 significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised.

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