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2 years ago

A fair number cube is rolled what is the probability that an odd number is rolled ?

Mathematics

2 answers:

maxonik [38]2 years ago

8 0

Answer:

B)1/2

Step-by-step explanation:

<u>Step 1: Find the odd numbers in a fair cube</u>

Total numbers in cube: 1,2,3,4,5,6 = 6

Odd numbers in a cube: 1,3,5 = 3

<u>Step 2: Apply the formula of probability</u>

<em>Probability = Number of favourable outcomes/Total number of outcomes</em>

Probability = 3/6

Probability = 1/2

B) 1/2 is the correct answer.

Licemer1 [7]2 years ago

7 0

Answer:

1/2

Step-by-step explanation:

-First you have to find the odd numbers in a fair cube

-Total numbers in cube would be: 1,2,3,4,5,6 = 6

-The odd numbers in the cube would be: 1,3,5 = 3

Step 2: Apply the formula of probability

-Probability = Number of favorable outcomes/Total number of outcomes

-Probability = 3/6 and 3/6 simplified is 1/2 so 1/2 is the answer

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Using the z-distribution, it is found that a sample of 171 should be selected.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$z = 1.96, \sigma = 40, M = 6$

Hence we solve for n to find the needed sample size.

$M = z\frac{\sigma}{\sqrt{n}}$

$6 = 1.96\frac{40}{\sqrt{n}}$

$6\sqrt{n} = 40 \times 1.96$

$\sqrt{n} = \frac{40 \times 1.96}{6}$

$(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2$

n = 170.7.

Rounding up, a sample of 171 should be selected.

More can be learned about the z-distribution at brainly.com/question/25890103

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