Ask question

Ask question

All categories

2 years ago

5

A bag contains red, green, and yellow marbles. The probability of selecting a red marble is 35%. What is the probability of not

selecting a red marble?

1 answer:

KonstantinChe [14]2 years ago

6 0

Probablity of selecting a red marble= 35%

Probablity of not selecting a red marble= 100%-35%

=65%

Hope it helps you. Please mark brainliest.

You might be interested in

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.6 ppm and standard de

Setler79 [48]

We assume that question b is asking for the distribution of $\\ \overline{x}$ , that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution $\\ X \sim N(8.6, 1.3)$ .

b. The distribution for the average amount of pollutants is $\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. $\\ P(z>-0.08) = 0.5319$ .

d. $\\ P(z>-0.47) = 0.6808$ .

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because <em>the sample was taken from a normal distribution</em>.

f. $\\ IQR = 0.2868$ ppm. $\\ Q1 = 8.4566$ ppm and $\\ Q3 = 8.7434$ ppm.

Step-by-step explanation:

First, we have all this information from the question:

The random variable here, X, is the number of pollutants that are found in waterways near large cities.
This variable is <em>normally distributed</em>, with parameters:
$\\ \mu = 8.6$ ppm.
$\\ \sigma = 1.3$ ppm.
There is a sample of size, $\\ n = 38$ taken from this normal distribution.

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with $\\ \mu = 8.6$ ppm and $\\ \sigma =1.3$ ppm or $\\ X \sim N(8.6, 1.3)$ .

b. What is the distribution of $\\ \overline{x}$ ?

The distribution for $\\ \overline{x}$ is $\\ N(\mu, \frac{\sigma}{\sqrt{n}})$ , i.e., the distribution for the sampling distribution of the means follows a normal distribution:

$\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not $\\ \overline{x}$ ). Then, we can use a <em>standardized value</em> or <em>z-score</em> so that we can consult the <em>standard normal table</em>.

$\\ z = \frac{x - \mu}{\sigma}$ [1]

x = 8.5 ppm and the question is about $\\ P(x>8.5)$ =?

Using [1]

$\\ z = \frac{8.5 - 8.6}{1.3}$

$\\ z = \frac{-0.1}{1.3}$

$\\ z = -0.07692 \approx -0.08$ (standard normal table has entries for two decimals places for z).

For $\\ z = -0.08$ , is $\\ P(z$ .

But, we are asked for $\\ P(z>-0.08) \approx P(x>8.5)$ .

$\\ P(z-0.08) = 1$

$\\ P(z>-0.08) = 1 - P(z$

$\\ P(z>-0.08) = 0.5319$

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is $\\ P(z>-0.08) = 0.5319$ .

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or $\\ P(\overline{x} > 8.5)$ ppm?

This random variable follows a standardized random variable normally distributed, i.e. $\\ Z \sim N(0, 1)$ :

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

$\\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ z = \frac{-0.1}{0.21088}$

$\\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47$

$\\ P(z$

Again, we are asked for $\\ P(z>-0.47)$ , then

$\\ P(z>-0.47) = 1 - P(z$

$\\ P(z>-0.47) = 1 - 0.3192$

$\\ P(z>-0.47) = 0.6808$

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is $\\ P(z>-0.47) = 0.6808$ .

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For $\\ P(z$ , $\\ z \approx -0.68$ , then, using [2]:

$\\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ (-0.68 *0.21088) + 8.6 = \overline{X}$

$\\ \overline{x} =8.4566$

$\\ Q1 = 8.4566$ ppm.

For Q3

$\\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ (0.68 *0.21088) + 8.6 = \overline{X}$

$\\ \overline{x} =8.7434$

$\\ Q3 = 8.7434$ ppm.

$\\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868$ ppm

Therefore, the IQR for the average of 38 cities is $\\ IQR = 0.2868$ ppm. $\\ Q1 = 8.4566$ ppm and $\\ Q3 = 8.7434$ ppm.

4 0

3 years ago

Layna can paint a 350-square foot room in 4 hours. It takes Bebe 7 hours to paint the same room. Which equation can be used to f

levacccp [35]

1/4+1/7=t
I am 99% sure this is correct. Good luck

4 0

2 years ago

What's the square footage of 12'3" * 18'4"

Olin [163]

Answer:

multiply length time width

4 0

3 years ago

Three boxes of cereal cost $9.57. How much do five boxes cost

Soloha48 [4]

the question is 9.57 × 5

Answer 47.85

7 0

2 years ago

Read 2 more answers

What's the addition property of 23 + 4 + 109 = 4 + 23 + 109

Grace [21]

Answer:

I am confused on this question too

Step-by-step explanation:

8 0

2 years ago