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Elden [556K]
2 years ago
9

Suppose that x has the density function f(x) = cx 0 < x < 2 identify c

Mathematics
1 answer:
ExtremeBDS [4]2 years ago
4 0

The value of C=1/3 for the probability density function f(x)=cx , 0<x<2.

<h3>What is probability density function?</h3>

The probability density function is a function of a continuous random variable, whose integral across an interval gives the probability that the value of the variable lies within the same interval.

Given that:

f(x)=cx\ \ at\ 0 < x < 2

Here we can see that the value of x is varies as 0,1,2

So at 0,1,2 the value of the function is

f(1)=c\times 1=c\\\\\\f(2)=c\times 2=2c\\\\

So the probability density function is given as:

\sum f(x)=f(1)+F(2)=1\\\\\sum f(x)=c+2c=1

3c=1\\\\c=\dfrac{1}{3}

Hence the value of C=1/3 for the probability density function f(x)=cx , 0<x<2.

To know more about probability density function follow

brainly.com/question/14214648

#SPJ4

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The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. (a) What is the probability
velikii [3]

Answer:

a) 89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200

b) No, because one of the requirements of the central limit theorem is a sample size of at least 30.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 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 = 10000, \sigma = 500

(a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200?

Here we have n = 40, s = \frac{500}{\sqrt{40}} = 79.06

This probability is the pvalue of Z when X = 10200 subtracted by the pvalue of Z when X = 9900. So

X = 10200

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

By the Central Limit Theorem

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

Z = \frac{10200 - 10000}{79.06}

Z = 2.53

Z = 2.53 has a pvalue of  0.9943.

X = 9900

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

Z = \frac{9900 - 10000}{79.06}

Z = -1.26

Z = -1.26 has a pvalue of  0.1038.

0.9943 - 0.1038 = 0.8905

89.05% probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200

(b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

No, because one of the requirements of the central limit theorem is a sample size of at least 30.

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