All categories

3 years ago

Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the 20th, 25th,65th , and 75th percentiles

SAT

1 answer:

Kay [80]3 years ago

7 0

Answer:

Percentile of 20th = 20

Percentile of 25th = 22.5

Percentile of 65th = 28

Percentile of 75th = 29

Explanation:

Sort the numbers in ascending order:

15,20,25,25,27,28,30,34

i = P/100 * N ( i index is the position of the number. Index 1 = 15, index 2 = 20, and so on)

Index of the 20th percentile:

20/100 * 8 = 1.6

Round 1.6 to the next integer = 2

The second number I=20

The 20th percentile is 20

Index of the 25th percentile:

25/100 * 8 = 2

Because 2 is an integer the percentile of 25th is the average of I=20 and I=25

(20+25)/2=22.5

The 25th percentile is 22.5

Index of the 65th percentile

i = 65/100 * 8 = 26.5 = 5.2

Round to the next integer = 6

The 65th percentile is 28

Index of the 75th percentile

I = 75/100 * 8 = 6

Since 6 is an integer (28+30)/2 = 29 percentile

You might be interested in

John bought 15 pines for $3500

Shalnov [3]

Answer:

Explanation:

If John needs to ensure a 40% profit, John needs to sell the pines at a price that would give 40% more money back than he spent.

Since John needs to make back all of the money he spent, plus another 40%, John needs to get back 140% of what he spent.

$140\% = 1.40$

John spent $3500.

140% of $3500 is $4900.

So, John must bring in $4900 (revenue) selling all 15 pines to achieve a 40% profit.

However, the question asks for the price of a single pine. To find the unit price of a single pine, divide the total revenue by 15.

$\frac{\$4900}{15pines}=\frac{\$326.\bar{6}}{pine}=\$326.66 \text{ per pine}$

6 0

2 years ago

the distribution of lengths of salmon from a certain river is approximately normal with standard deviation 3.5 inches. if 10 per

juin [17]

Complete question;

<em>The distribution of lengths of salmon from a certain river is approximately normal with a standard deviation of 3.5 inches. If 10 percent of salmon are longer than 30 inches, which of the following is closest to the mean of the distribution? 26 inches A 28 inches B 30 inches C 33 inches D 34 inches</em>

Option B is correct. The value that is closest to the mean of the distribution is 30inches.

The formula for calculating the z-score is expressed as:

$z=\frac{ x-\mu}{\sigma}$

Given the following parameters

$\sigma = 3.5\\x = 26in$

If 10 percent of salmon are longer than 30 inches, then:

P(x>30) = 0.1

Using the z table to get the value corresponding to the mean area 0.1.

The required z-score will be -1.285

Substitute the resulting parameters into the formula to get the mean of the distribution.

$z=\frac{x-\mu}{\sigma}\\-1.285=\frac{26-\mu}{3.5}\\-1.285 \times 3.5 = 26 - \mu\\ -4.4975 = 26 - \mu\\\mu = 26 + 4.4975\\\mu = 30.4975\\\mu \approx 30in$

Hence the value that is closest to the mean of the distribution is 30inches.

Learn more here: brainly.com/question/15295437

3 0

2 years ago

The writer wants to add an additional body paragraph to support the passage's argument. Which of the following claims would make

Tamiku [17]

Answer:

E - Long-term missions would also create social and psychological issues, such as extreme isolation and loneliness, that may have a negative impact on space travelers.

Explanation:

just did the test.

8 0

3 years ago

Read 2 more answers

Complete the following equations (note that the equations are not balanced). Use the activity series if necessary. An activity s

balandron [24]

Answer:

PbSO4 + MgCl2

Explanation:

got it right on edge

5 0