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sergij07 [2.7K]
3 years ago
11

suppose X and Y are independent random variables, both with normal distributions. If X has a mean of 45 with a standard deviatio

n of 4 and Y has a mean of 35 with a standard deviation of 3, what is the probability that a randomly generated value of X is greater than a randomly generated value of Y
Mathematics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu_X = 45, \sigma_X = 4, \mu_Y = 35, \sigma_Y = 3

What is the probability that a randomly generated value of X is greater than a randomly generated value of Y

This means that the subtraction of X by Y has to be positive.

When we subtract two normal variables, the mean is the subtraction of their means, and the standard deviation is the square root of the sum of their variances. So

\mu = \mu_X - \mu_Y = 45 - 35 = 0

\sigma = \sqrt{\sigma_X^2+\sigma_Y^2} = \sqrt{25} = 5

We want to find P(X > 0), that is, 1 subtracted by the pvalue of Z when X = 0. So

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

Z = \frac{0 - 10}{5}

Z = -2

Z = -2 has a pvalue of 0.0228

1 - 0.0228 = 0.9772

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

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Answer:

(-5, 6) is not a solution.

Step-by-step explanation:

6=2(-5)+4

6=-10+4

-10+4=-6, not 6.

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She would need to work for 35 hours.

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vesna_86 [32]

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What is this question asking

Step-by-step explanation:

3 0
3 years ago
A company claims that less than 10% of adults own a smart watch. You want to test this claim, and you find that in a random samp
Nezavi [6.7K]

Answer:

Test statistic is 0.67

Critical value is -2.33

Step-by-step explanation:

Consider the provided information.

The formula for testing a proportion is based on the z statistic.

z=\frac{\hat p-p_0}{\sqrt{p_0\frac{1-p_0}{n}}}

Were \hat p is sample proportion.

p_0 hypothesized proportion and n is the smaple space,

Random sample of 100 adults, 12% say that they own a smart watch.

A company claims that less than 10% of adults own a smart watch.

Therefore, n = 100  \hat p = 0.12 , P_0 = 0.10

1 - P_0 = 1 - 0.10 = 0.90

Substitute the respective values in the above formula.

z=\frac{0.12-0.10}{\sqrt{0.10\frac{0.90}{100}}}

z\approx 0.67

Hence, test statistic = 0.67

This is the left tailed test.

Now using the table the P value is:

P(z < 0.667) = 0.7476

P-value = 0.7476

\alpha = 0.01

Here,  P-value > α therefore, we are fail to reject the null hypothesis.

Z_{\alpha}= Z_{0.01} = -2.33

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7 0
3 years ago
The expression can be used to predict the height of a plant in days. Where is the rate in centimeters that that plant grows per
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Answer:

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Step-by-step explanation:

x = days

f(x) = 1.5x + 30

f(5) = 1.5(5) + 30

f(5) = 7.5 + 30

f(5) = 37.5

5 0
3 years ago
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