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Marrrta [24]
2 years ago
5

The data set below represents the ages of 36 executives. find the percentile that corresponds to an age of 4141 years old. 2828

2929 2929 3232 3232 3333 3434 3434 3434 3434 3737 3737 3838 4141 4141 4242 4545 4545 4747 4747 4747 4848 5050 5151 5353 5656 5656 5656 6161 6161 6262 6363 6464 6464 6565 6666
Mathematics
1 answer:
solong [7]2 years ago
8 0

Answer:

37th percentile.

Step-by-step explanation:

We have been given a data set that represents the ages of 36 executives. We are asked to find the percentile that corresponds to an age of 41 years.

28, 29, 29, 32, 32, 33, 34, 34, 34, 34, 37, 37, 38, 41, 41, 42, 45, 45, 47, 47, 47, 48, 50, 51, 53, 56, 56, 56, 61, 61, 62, 63, 64, 64, 65, 66.

Let us count the number of data points below and at 41.

We can see that the number of data points at and below 41 is 13.

We will use percentile formula to solve our given problem.

\text{Percentile rank of x}=\frac{\text{Number of values below x}}{\text{Total number of data points}}\times 100

\text{Percentile rank of 41}=\frac{13}{36}\times 100

\text{Percentile rank of 41}=0.361111\times 100

\text{Percentile rank of 41}=36.11\approx 37

Therefore, the percentile rank that corresponds to age of 41 years old is 37th percentile.  

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Part B: When there are equal number of participants in both campuses, stratification by campus would give a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender.  Another method to ensure that stratification by campus gives a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender is to ensure that the sample is proportional to the proportion of each campus to the whole population or workforce.

Step-by-step explanation:

A Convenience Sampling technique is a non-probability (non-random) sampling method and the participants are selected based on availability (early attendees).  The early attendees might be different from the late attendees in characteristics such as age, sex, etc.  Therefore, sampling biases are present.  All non-probability sampling methods are prone to volunteer bias.

Stratified sampling  is more accurate and representative of the population.  It reduces sampling bias.  The difficulty arises in choosing the characteristic to stratify by.

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Step-by-step explanation:

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Welcome
5 0
2 years ago
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Find the values of k for which the line y=1-2kx does not meet the curve y=9x²-(3k+1)x+5
Lilit [14]

Let's equate the two given functions and attempt to solve for x:

y = 1 -2kx = y = 9x^2 -(3k+1)x + 5

Eliminating y, 1 -2kx = 9x^2 -(3k+1)x + 5

Rearranging terms in descending order by powers of x:

0 = 9x^2 - (3k+1)x + 2kx + 5 - 1 , or

0 = 9x^2 - kx - x + 4

This is a quadratic equation with coefficients a = 9, b = -(k+1) and c = 4.

For certain k, not yet known, solutions exist. Solutions here implies points at which the two curves intersect.

k+1 plus or minus sqrt( [-(k+1)]^2 - 4(9)(4) )

x = -----------------------------------------------------------------

2(9)

The discriminant is k^2 + 2k + 1 - 144, or k^2 + 2k - 143.

If the discriminant is > 0, there are two real, unequal roots. We don't want this, since we're interested in finding k value(s) for which there's no solution.

If the discr. is = 0, there are two real, equal roots. Again, we don't want this.

If the discr. is < 0, there are no real roots. This is the case that interests us.

So our final task is to determine the k values for which the discr. is < 0:

Determine the k value(s) for which the discriminant, k^2 + 2k - 143, is 0.

This k^2 + 2k - 143 factors as follows: (k-11)(k+13), and when set = to 0, results in k: {-13,11}.

Set up intervals on the number line: (-infinity, - 13), (-13, 11) and (11, infinity).

Choosing a test number from each interval, determine the interval or intervals on which the discriminant is negative:

Case 1: k = -15; the discriminant (k^2 + 2k - 143) is (-15)^2 + 2(-15) - 143 = +52. Reject this interval

Case 2: k = 0; the discriminant is then 0 + 0 - 143 (negative); thus, the discriminant is negative on the interval (-13,11).

Case 3: k = 20; the discriminant is positive. Reject this interval.

Summary: The curves do not intersect on the interval (-13,11).

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