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3 years ago

9

Rylan is calculating the standard deviation of a data set that has 9 values. He determines that the sum of the squared deviation

s is 316. What is the standard deviation of the data set? Round the answer to the nearest tenth

2 answers:

stepladder [879]3 years ago

6 0

The formula for the sum of the squares of the deviations is
SS = s² (N -1)

We are given
SS = 316
N = 9

Substituting
316 = s² (9 - 1)
Solving for s
s = 6.28

The standard deviation is 6.28

olga_2 [115]3 years ago

5 0

Answer:

6.3

Step-by-step explanation:

We have been given that Rylan is calculating the standard deviation of a data set that has 9 values. He determines that the sum of the squared deviations is 316.

We will use the formula $\text{Standard deviation}=\sqrt{\frac{\text{Sum of squares}}{n-1}}$ , where n represents the number of data points in a data set.

Upon substituting our given values in above formula we will get,

$\text{Standard deviation}=\sqrt{\frac{316}{9-1}}$

$\text{Standard deviation}=\sqrt{\frac{316}{8}}$

$\text{Standard deviation}=\sqrt{39.5}$

$\text{Standard deviation}=6.284902544988\approx 6.3$

Therefore, the standard deviation of our given data set is 6.3.

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The cost of producing n necklaces is p(n) = 15n + 60. The necklaces cost $30, which can be represented by c(n) = 30n. For how ma

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Step-by-step explanation:

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3 years ago

If you draw a card with a value of 4 or less from a standard deck of cards, I will pay you 165$ if not your pay me $45. (Aces ar

jekas [21]

Answer: The required expected value is $3.46.

Step-by-step explanation:

Since we have given that

Number of value of 4 or less are

2,3,4

So, there are 12 in numbers.

So, probability would be

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So, the expected value would be

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3 years ago

The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab

kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.

The mean of the approximated distribution of X is:

μ = λ = 250

The standard deviation of the approximated distribution of X is:

σ = √λ = √250 = 15.8114

Thus, $X\sim N(250, 250)$

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

$P(235$

$=P(-0.95$

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.28$

Thus, the value of $P(48$ .

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

$P(48$

$=P(-0.40$

Thus, the value of $P(48$ .

(d)

Let the sample size be <em>n</em>.

$P(48$

$0.95=P(-z$

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

$z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241$

Thus, the sample selected must be of size 240.

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