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Leto [7]
3 years ago
13

The effect of changing the standard deviation. (a) Sketch a Normal curve that has mean 10 and standard deviation 3. (b) On the s

ame x axis, sketch a Normal curve that has mean 10 and standard deviation 1. (c) How does the Normal curve change when the standard deviation is varied but the mean stays the same?

Mathematics
1 answer:
aalyn [17]3 years ago
4 0

Answer:

See explanation below.

Step-by-step explanation:

Part a

We can use the following R code:

> x <- seq(0, 20, length=1000)  # Generate 1000 sample between 0 and 20

> y <- dnorm(x, mean=10, sd=3)   # Generate the probability density values for each value in the vector x defined with mean = 10 and sd= 3

Part b

> y1<-dnorm(x, mean=10,sd=1)  # Generate the probability density values for the normal distribution with mean 10 and sd =1

> plot(x, y1, type="l", lwd=1, col="green")   # Plot the density function for the normal distribution with mean =10 and sd =1

We can see on the figure attached that the green line represent X \sim N(\mu = 10, \sigma=1)

> lines(x,y, col="red")

And the red line would represent the Y \sim N (\mu = 10, \sigma=3)

Part c

As we can see when we reduce the deviation the spread of the distribution is lower, and we have most of the values near to the mean, and in the other case when the deviation increases we have more spread on the distribution with the values more dispersed in the x axis.

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52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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Normal probability distribution:

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The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

Z = \frac{30732 - 30393}{472.81}

Z = 0.72

Z = 0.72 has a pvalue of 0.7642.

X = 30054

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

Z = \frac{30054 - 30393}{472.81}

Z = -0.72

Z = -0.72 has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

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