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Chapter 4: Q133E (page 276)

4.133 Suppose xis a random variable best described by a uniform

probability distribution with c= 20 and d= 45.

a. Find f(x)

b. Find the mean and standard deviation of x.

c. Graph f (x) and locate 渭and the interval 渭卤2蟽onthe graph. Note that the probability that xassumes avalue within the interval 渭卤2蟽is equal to 1.

Short Answer

Expert verified

a. The probability density function is

f(x)=0.0420鈮�x鈮�450;otherwise

b. The mean is 32.5 and the standard deviation is 7.2169

c. The lower limit is 18.0662 and the upper limit is 46.9338

Step by step solution

01

Given Information

Here x is a uniform random variable with parameters c=20 and d=45.

02

Finding the f (x)

a.

The probability density function random variable x is given by

f(x)=1d-c;c<x<d

Here, c=20 and d=45

So the pdf of x is:

f(x)=145-20=125=0.04

Thus, f (x0 = 0.04 ; 20 < x <78C8 45

03

Finding the mean and standard deviation of x.

b.

The mean of the random variable x is given by,

渭=c+d2=20+452=652=32.5

The standard deviation of x is given by,

蟽=d-c12=45-2012=2523=7.2169

Thus, the mean 渭=32.5and standard deviation 蟽=7.2169.

04

The Graph

c.

Here, the 2蟽limit is given by,

渭-2蟽=32.5-2脳7.2169=18.0662渭+2蟽=32.5+2脳7.2169=46.9338

Here, the interval limits are outside of the actual range of random variable x.

The following graph depicts the relevant situation.

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Most popular questions from this chapter

Establishing tolerance limits. The tolerance limits for a product's quality characteristic (e.g., length, weight, or strength) are the minimum or maximum values at which the product will operate properly. Tolerance limits are set by the engineering design function of the manufacturing operation (Total Quality Management, Vol. 11, 2000). The tensile strength of a particular metal part can be characterized as being normally distributed with a mean of 25 pounds and a standard deviation of 2 pounds. The part's upper and lower tolerance limits are 30 pounds and 21 pounds, respectively. A part that falls within the tolerance limits results in a profit of \(10. A part that falls below the lower tolerance limit costs the company \)2; a part that falls above the upper tolerance limit costs the company $1. Find the company鈥檚 expected profit per metal part produced.

If x is a binomial random variable, use Table I in Appendix D to find the following probabilities:

a.for n = 10, p = .4

b.for n = 15, p = .6

c.for n = 5, p = .1

d.for n = 25, p = .7

e.for n = 15, p = .9

f.for n = 20, p = .2

Consider the discrete probability distribution shown here:

  1. Find 渭=贰(x).
  2. Find 蟽=E[(x鈭�渭)2]
  3. Find the probability that the value of x falls within one standard deviation of the mean. Compare this result to the Empirical Rule.

The binomial probability distribution is a family of probability distributions with every single distribution depending on the values of n and p. Assume that x is a binomial random variable with n = 4.

  1. Determine a value of p such that the probability distribution of x is symmetric.
  2. Determine a value of p such that the probability distribution of x is skewed to the right.
  3. Determine a value of p such that the probability distribution of x is skewed to the left.
  4. Graph each of the binomial distributions you obtained in parts a, b, and c. Locate the mean for each distribution on its graph.\
  5. In general, for what values of p will a binomial distribution be symmetric? Skewed to the right? Skewed to the left?

If x is a binomial random variable, compute for each of the following cases:

  1. n = 4, x = 2, p = .2
  2. n = 3, x = 0, p = .7
  3. n = 5, x = 3, p = .1
  4. n = 3, x = 1, p = .9
  5. n = 3, x = 1, p = .3
  6. n = 4, x = 2, p = .6
See all solutions

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