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2 years ago

11

Shawnte earned a score of 42 points. There were 19 competitors who received a lower score than Shawnte and 5 competitors who ear

ned a higher score. What was Shawnte's percentile rank in the bodybuilding competition?

1 answer:

Natasha_Volkova [10]2 years ago

5 0

Answer: 80th percentile

Step-by-step explanation:

There were 19 competitors below him and 5 competitors above him.

Total number of competitors is:

= 19 + 5 + Shawnte

= 25 competitors

Out of these 25 competitors, Shawnte took the 20th position.

Percentile is:

= 20/25 * 100

= 80th percentile

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An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

(can someone help??) <br> −12x+8=6x−12−13/2x

worty [1.4K]

Answer:

t

Step-by-step explanation:

h

4 0

2 years ago

Suppose that prices of recently sold homes in one neighborhood have a mean of $265,000 with a standard deviation of $9300. Using

Hatshy [7]

Answer:

Range = (237100, 292900)

Step-by-step explanation:

Using Chebyshevs Inequality:

$P(|X - \mu | \le k \sigma )\ge 1 -\dfrac{1}{k^2}= 0.889$

$1 -\dfrac{1}{k^2}= 0.889$

$\dfrac{1}{k^2}= 1- 0.889$

$\dfrac{1}{k^2}=0.111$

$k = \sqrt{\dfrac{1}{0.111}}$

$k \simeq 3$

Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ ± kσ

where;

μ = 265000

σ = 9300

Range = 265000 ± 3(9300)

Range = 265000 ± 27900

Range = (265000 - 27900, 265000 + 27900)

Range = (237100, 292900)

3 0

2 years ago

Fine the length of the hypotenuse of a Right triangle with legs pf 20cm and21cm

Arisa [49]

Hypotenuse² = perpendicular² + base²

H² = 20²+21²

H²= 400+441 = 841

H = √841

H = 29 cm

3 0

2 years ago

5. Amber buys bottles of lemonade and a bag of cookies at the store. She pays a total of $9.50. The bag of cookies cost $3.25. I

Anika [276]

Answer:

each bottle was 1.25

Step-by-step explanation:

subtract 3.25 from 9.50 and then divide that answer by 5 for the 5 bottles of lemonade

3 0

2 years ago

Read 2 more answers