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2 years ago

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A bowl has 3 red beads and 7 blue beads. Two beads are selected at random one at a time. First a red bead is drawn and placed on

a table. Next a blue bead is drawn.
What is the probability of this event?

Drag and drop the answers into the boxes to complete the sentences.

1 answer:

IceJOKER [234]2 years ago

7 0

The probability that first a red bead is drawn and next a blue bead is drawn is 7/30,and both events are dependent

<h3>How to determine the probability?</h3>

The distribution of the beads are:

Red = 3

Blue = 7

Total = 10

The probability of selecting the red bead first is:

P(Red) = 3/10

When the red bead is selected, the number of beads becomes 9

So, the probability of selecting a blue bead is

P(Blue) = 7/9

The probability of the event is then calculated using:

P = 3/10 * 7/9

Evaluate

P = 7/30

Hence, the probability of the event is 7/30

Read more about probability at:

brainly.com/question/25638875

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2 years ago

A certain industrial process yields a large number of steel cylinder whose lengths are distributed normally with mean 3.25 inche

Taya2010 [7]

Answer:

$Y = X_1 + X_2$

If we find the expected value of Y we got:

$E(Y) = E(X_1) + E(X_2) = 3.25 + 3.25 =6.5$

And for the variance we assume that we have independent random variables for X1 and X2 so then we have:

$Var(Y) = Var(X_1) + Var(X_2) = \sigma^2 + \sigma^2 = 2 \sigma^2 = 2*(0.05^2) = 0.005$

And the deviation $Sd(Y) = \sqrt{0.005}=0.0707$

So then our random variable Y have the following distribution:

$Y \sim N (\mu =6,5 , \sigma = 0.0707)$

And we want to find this probability:

$P(Y$

And we can use the z score formula given by:

$z=\frac{y -\mu_y}{\sigma_y}$

And if we use the z score formula we got:

$P(Y$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.25,0.05)$

Where $\mu=3.25$ and $\sigma=0.05$

We slect two observations $X_1, X_2$

And let denote the combined lenght as:

$Y = X_1 + X_2$

If we find the expected value of Y we got:

$E(Y) = E(X_1) + E(X_2) = 3.25 + 3.25 =6.5$

And for the variance we assume that we have independent random variables for X1 and X2 so then we have:

$Var(Y) = Var(X_1) + Var(X_2) = \sigma^2 + \sigma^2 = 2 \sigma^2 = 2*(0.05^2) = 0.005$

And the deviation $Sd(Y) = \sqrt{0.005}=0.0707$

So then our random variable Y have the following distribution:

$Y \sim N (\mu =6,5 , \sigma = 0.0707)$

And we want to find this probability:

$P(Y$

And we can use the z score formula given by:

$z=\frac{y -\mu_y}{\sigma_y}$

And if we use the z score formula we got:

$P(Y$

3 0

2 years ago

In a previous year, 58% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different tod

Gemiola [76]

Answer:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$

Since is a bilateral test the p value would be given by:

$p_v =2*P(z$

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58

Step-by-step explanation:

Information given

n=600 represent the random sample selcted

X=339 represent the number of females aged 15 and older that living alone

$\hat p=\frac{339}{600}=0.565$ estimated proportion of females aged 15 and older that living alone

$p_o=0.58$ is the value that we want to check

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_$ represent the p value

Sytem of hypothesis

We want to check if the true proportion females aged 15 and older that living alone is significantly different from 0.58.:

Null hypothesis: $p=0.58$

Alternative hypothesis: $p \neq 0.58$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$

Since is a bilateral test the p value would be given by:

$p_v =2*P(z$

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58

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3 years ago

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Answer:

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Step-by-step explanation:

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