Is it possible capture the correlation between continuous and categorical variable?

1 answer:

7 0

It is possible to<span> capture the correlation (or lack thereof) between continuous and categorical variable using Analysis of Covariance (</span>ANCOVA) technique to capture association among continuous and categorical variables.

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Find a constant c for which p(z ≥ c.= 0.1587. round the answer to two decimal places.

Vikki [24]

What you see in the z score table is P(z< a constant), for P( z≥ a number), subtract the probability from 1:
1-0.1587=0.8413
on the z score table, you will see that p(z<1.00)=0.8413
so c=1.00

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2 years ago

The points in the table lie on a line. Find the slope of the line.

wolverine [178]

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Y=5x+2

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3 years ago

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Answer:

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A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of

Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .