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Mila [183]
1 year ago
8

A spinner has five congruent sections, one each of blue, green, red, orange, and yellow. Yuri spins the spinner 10 times and rec

ords his results in the table. A 2-column table has 5 rows. The first column is labeled Color with entries blue, green, red, orange, yellow. The second column is labeled Number with entries 1, 2, 0, 4, 3. Which statements are true about Yuri’s experiment? Select three options. The theoretical probability of spinning any one of the five colors is 20%. The experimental probability of spinning blue is One-fifth. The theoretical probability of spinning green is equal to the experimental probability of spinning green. The experimental probability of spinning yellow is less than the theoretical probability of spinning yellow. If Yuri spins the spinner 600 more times and records results, the experimental probability of spinning orange will get closer to the theoretical probability of spinning orange.
Mathematics
1 answer:
Burka [1]1 year ago
3 0

Answer:

S = {AB, AC, BC}

Step-by-step explanation:

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A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr
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Answer:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:P(-1.5 \leq Z \leq 1.5) = P(zStep-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

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 variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(50,12)  

Where \mu=50 and \sigma=12

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required like this:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:

P(-1.5 \leq Z \leq 1.5) = P(z

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