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NikAS [45]
3 years ago
10

An oil company is interested in estimating the true proportion of female truck drivers based in five southern states. A statisti

cian hired by the oil company must determine the sample size needed in order to make the estimate accurate to within 2% of the true proportion with 99% confidence. What is the minimum number of truck drivers that the statistician should sample in these southern states in order to achieve the desired accuracy?
Mathematics
1 answer:
AlekseyPX3 years ago
7 0

Answer: n = 2401

Step-by-step explanation:

Given;

Confidence level = 2% - 99%

n = ? ( which is the sample size is unknown ).

Solution:

Where;

n = [z/E]^2*pq

Since no known value for ( p ) estimate is given, the "least biased" estimate is p = 1/2

Substituting the given data into the formula.

n = [1.96/0.02]^2(1/2)(1/2)

n = 2401

The minimum number of truck drivers the statistician needs to sample for an accurate result is 2401

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zmey [24]

Answer:

\large\boxed{Q1.\ \sin\theta=\dfrac{40}{41}}\\\boxed{Q2.\ 30^o}

Step-by-step explanation:

Q1.

\sin\theta=\dfrac{y}{r}

\text{We have}\\\\r=1\to(\text{radius of the unit circle}),\\\\y=\dfrac{40}{41}\to\text{from the given point}\ \left(\dfrac{9}{41},\ \dfrac{40}{41}\right)\\\\\text{Substitute:}\\\\\sin\theta=\dfrac{\frac{40}{41}}{1}=\dfrac{40}{41}

Q2.

<em>look at the picture</em>

210° is in III Quadrant. Therefore the reference angle for θ is:

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3 years ago
A 2-column table with 10 rows. Column 1 is labeled Classes with entries Hall, Benny, Leggo, Talle, Flower, Gomez, Range, Book, T
Alex787 [66]

Answer:

range: 20

median: 38

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Interquartile range: 13.5

Which value was affected the most: range

8 0
3 years ago
Order the numbers from least to greatest<br> 6.23<br> 6.023<br> 0.623<br> 62.30
mrs_skeptik [129]

Answer:

0.623, 6.023, 6.23, 62.30

Step-by-step explanation:

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3 years ago
According to a survey by the Administrative Management Society, one-half of U.S. companies give employees 4 weeks of vacation af
andreyandreev [35.5K]

Answer:

a

P(2\leq X\leq 5)=0.875

b

P(X

Step-by-step explanation:

Let the random X variable representing the 6 companies that give 4 weeks of vacation after 15 years of employment:

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#Probability that the number of companies that give vacation is anywhere from 2 to 5:

We use equation 1;

P(2\leq X\leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\={6 \choose 2}(0.5)^6+{6 \choose 3}(0.5)^6{6 \choose 4}(0.5)^6+{6 \choose 5}(0.5)^6\\\\=0.0156(15+20+15+6)\\\\=0.875

Hence the probability that between 2 and 5 companies give vacation is 0.875

b. The probability that fewer than 3 companies give vacation is calculated as:

From equation one we get:

P(X

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3 years ago
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Step-by-step explanation:

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3 years ago
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