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2 years ago

Problem 2. A random variable has the following density function:

Mathematics

1 answer:

kolbaska11 [484]2 years ago

6 0

If f(x) is a density function, then its integral over the entire real line should be 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_0^2(1-0.5x)\,\mathrm dx=(x-0.25x^2)\bigg|_0^2=2-1=1$

so yes, f(x) is indeed a density function.

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Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results.Listed bel

timama [110]

Answer:

D.No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.

Step-by-step explanation:

(a) Range is the difference between the smallest and largest observation.

Here Smallest observation = 0.63

and Largest observation = 1.48

⇒ Range = 0.85

(b) Standard Deviation is calculate by,

$Standard deviation(\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\bar{x})^{2}} }$

where, $\bar{x}$ is mean of the observation.

Here, Mean = 0.988

⇒ Standard Deviation = 0.313

(c) Variance is the square of Standard deviation.

Thus, Variance = (Standard Deviation)² = 0.098

(d) Here last option(D) is true i.e. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.

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3 years ago

Help with two step equations with fractions......... 2/3x=6

pashok25 [27]

Multiply 2/3 by the reciprocal, 3/2, on both sides.

3 0

3 years ago

Read 2 more answers

A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A st

Anon25 [30]

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

Step-by-step explanation:

Denote the events as follows:

X = an US adult who does not uses social media.

Y = an US adult between the ages 18 and 29.

Z = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (X')) = 1 - P (US adult so not use social media)

$=1-P(X)\\=1-0.35\\=0.65$

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (Y)) = 1 - P (Adults 30 or above)

$=1-P(Z)\\=1-0.78\\=0.22$

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

$=0.22+0.65-0.672\\=0.198$

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

6 0

3 years ago

Can u please answer this sme one

nata0808 [166]

Answer:

Samir wins by 20 seconds

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The number of flaws per square yard in a type of carpet material varies with mean 1.4 flaws per square yard and standard deviati

oksian1 [2.3K]

According to the Central Limit Theorem, the distribution of the sample means is approximately normal, with the mean equal to the population mean (1.4 flaws per square yard) and standard deviation given by:
$\frac{\sigma}{ \sqrt{n} }=\frac{1.2}{ \sqrt{178} }=0.09$
The z-score for 1.5 flaws per square yard is:
$z=\frac{1.5-1.4}{0.09}=1.11$
The cumulative probability for a z-score of 1.11 is 0.8665. Therefore the probability that the mean number of flaws exceeds 1.5 per square yard is
1 - 0.8665 = 0.1335.

6 0

3 years ago