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gladu [14]
2 years ago
15

Given the graph below, find GH.

Mathematics
1 answer:
Alex777 [14]2 years ago
3 0

Answer:

GH = 6.3 (nearest tenth)

Step-by-step explanation:

Given:

G(-2, -3)

H(0, 3)

Required:

Distance between point G and H = GH

Solution:

Distance between G(-2, -3) and H(0, 3):

GH = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

G(-2, -3) = (x_1, y_1)

H(0, 3) = (x_2, y_2)

GH = \sqrt{(0 - (-2))^2 + (3 - (-3))^2}

GH = \sqrt{(2)^2 + (6)^2}

GH = \sqrt{4 + 36} = \sqrt{40}

GH = 6.3 (nearest tenth)

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The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm
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Answer:

a) P(12.99 ≤ X ≤ 13.01) = 0.3840

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Step-by-step explanation:

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

Normal probability distribution

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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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 13, \sigma = 0.08

(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.

Here we have n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02

This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.

X = 13.01

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

By the Central Limit Theorem

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

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

X = 12.99

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

Z = \frac{12.99 - 13}{0.02}

Z = -0.5

Z = -0.5 has a pvalue of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?

P(X ≥ 13.01) =

This is 1 subtracted by the pvalue of Z when X = 13.01. So

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

Z = \frac{13.01 - 13}{0.02}

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1 - 0.6915 = 0.3075

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