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lutik1710
3 days ago
13

A sample of size 200 will be taken at random from an infinite population. given that the population proportion is 0.60, the prob

ability that the sample proportion will be greater than 0.58 is
Mathematics
1 answer:
Leona [4.1K]3 days ago
4 0

Let p represent the proportion of the population. <span>
Given p=0.60, n=200 we need to calculate P(^p<0.58).
</span>
According to the rule of thumb, since n*p = 200*0.60 and n*(1-p)= 200*(1-0.60) = 80, both values exceed 5, confirming that n is large enough for the sampling distribution of sample proportion-^p to comply with the z standard normal distribution. The mean for the sampling distribution will be U^p = p = 0.60, and the standard deviation, δ^p = √[p*(1-p)/n] = √[0.60*(1-0.60)/200] = √0.0012.
Thus, the probability that the sample proportion is below 0.58
= P(^p<0.58)
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0...
= P(z<-0.58)
= P(z<0) - P(-0.58<z<0)
= 0.5 - 0.2190
= 0.281
<span>Thus, the likelihood that the sample proportion is smaller than 0.58 stands at 0.281 or 28.1%.</span>

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I have included a screenshot of the entire question along with its accompanying diagram.

Answer:
∠1 = 163°

Explanation:
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I hope this clarifies things!:)

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4 days ago
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Zina [3893]

Answer:

C. The mean daily salary is not above $350 per day

Step-by-step explanation:

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Y = a + bX

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4 days ago
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Now, converting 4 hours into seconds.

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9 days ago
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9 0
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Response:

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4 0
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