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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

Short Answer

Expert verified
  1. P(x = 2)=0.121.
  2. P(x⩽5)=0.034.
  3. P(x > 1)=0.001.
  4. role="math" localid="1655726308157" p(x<10)=0.000.
  5. P(x⩾10)=0.998.
  6. P(x = 2)=0.137.

Step by step solution

01

Formula for calculating  P(x = 2)

a.

Considering the specific value of n (number of trials), p(number of successes), and x(binomial random variable), the formula for calculating Pl(X=2)is shown below.

Px=2=Px⩽2-Px⩽1

By subtracting Px⩽1from Px⩽2, the value of Px = 2can be found.

02

Calculation of P(x = 2)

The calculation of is shown below.

Px=2=Px⩽2-Px⩽1=0.167-0.046=0.121

The computed value of Px = 2is 0.121.

03

Formula for calculating  P(x≤5)

b.

Considering the specific value of n as 15 and that of p as 0.6, the formula for calculatingP(x≤5) is shown below.

Px≤5=∑a5px

04

Determining the value of  Px⩽5

InTable 1, when n is 15 and p is 0.6,the value along the row containing k as 5 is shown below.

Px⩽5=∑i=05px=0.034

The value ofP(x⩽5)is 0.343.

05

Formula for calculating Px > 1

By considering the specific value of n as 5 and that of p as 0.1, the method for calculating Px > 1is shown below.

Px>1=∑i=25px

06

Computing the value of  P(x > 1)

InTable 1, when n is 5 and p is 0.1, the value ofPx⩽1 must be subtracted from 1 to get the value of Px > 1:

Px>1=∑i=25px=1-Px⩽1=1-0.999=0.001

The computed value ofPx > 1 is 0.001.

07

 Step 7: Formula for calculating px<10

By considering the specific value of n as 25 and that of p as 0.7, the method for calculating px<10is shown below.

Px<10=∑i=09px=Px⩽9

08

Computing the value of  Px≤10 

For calculating px≤10Px⩽9,will be considered, and the calculation is shown below.

Px<10=Px⩽9=0.000

The value ofPx > 1will therefore be 0.000.

09

Formula for calculating Px⩾10

By considering the specific value of n as 15 and that of p as 0.9, the method for calculating Px⩾10is shown below.

Px⩾10=∑i=1015px=1-Px⩽9

10

Computing the value of Px⩾10

For calculatingPx⩾10P(x⩽9) ,must be subtracted from 1.Thecalculation is shown below.

Px⩾10=∑i=1015px=1-Px⩽9=1-0.002=0.998

The value of P(x⩾10)will therefore be 0.998.

11

Formula for calculating P(x = 2)

Considering the specific value of n as 20 and that of p as 0.2, the formula for calculatingPx = 2is shown below.

Px=2=Px⩽2-Px⩽1

By subtracting the table value Px⩽1from that of Px⩽2, the value of P(x = 2)can be found.

12

Calculation ofP(x = 2)

The calculation of Pl(X=2)with the above formula is shown below.

role="math" localid="1655726498573" Px=2=Px⩽2-Px⩽1=0.206-0.069=0.137Px=2=Px⩽2-Px⩽1=0.206-0.069=0.137

The computed value ofP(x = 2)is 0.137.

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Most popular questions from this chapter

Find the area under the standard normal probability distribution between the following pairs of z-scores:

a)z=0andz=2.00

b)z=0andz=3

c)z=0andz=1.5

d)z=0andz=.80


Industrial filling process. The characteristics of an industrialfilling process in which an expensive liquid is injectedinto a container were investigated in the Journal of QualityTechnology(July 1999). The quantity injected per containeris approximately normally distributed with mean 10

units and standard deviation .2 units. Each unit of fill costs\(20 per unit. If a container contains less than 10 units (i.e.,is underfilled), it must be reprocessed at a cost of \)10. A properly filled container sells for $230.

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Under these conditions, what is the expected profit percontainer?

Suppose x is a binomial random variable with p = .4 and n = 25.

a. Would it be appropriate to approximate the probability distribution of x with a normal distribution? Explain.

b. Assuming that a normal distribution provides an adequate approximation to the distribution of x, what are the mean and variance of the approximating normal distribution?

c. Use Table I in Appendix D to find the exact value of P(x≥9).

d. Use the normal approximation to find P(x≥9).

Checkout lanes at a supermarket. A team of consultants working for a large national supermarket chain based in the New York metropolitan area developed a statistical model for predicting the annual sales of potential new store locations. Part of their analysis involved identifying variables that influence store sales, such as the size of the store (in square feet), the size of the surrounding population, and the number of checkout lanes. They surveyed 52 supermarkets in a particular region of the country and constructed the relative frequency distribution shown below to describe the number of checkout lanes per store, x.

a. Why do the relative frequencies in the table represent the approximate probabilities of a randomly selected supermarket having x number of checkout lanes?

b. FindE(x) and interpret its value in the context of the problem.

c. Find the standard deviation of x.

d. According to Chebyshev’s Rule (Chapter 2, p. 106), what percentage of supermarkets would be expected to fall withinμ±σ? withinμ±2σ?

e. What is the actual number of supermarkets that fall within? ? Compare your answers with those of part d. Are the answers consistent?

Suppose xis a binomial random variable with n= 20 and

p= .7.

a. FindP(x=14).

b. FindP(x≤12).

c. FindP(x>12).

d. FindP(9≤x≤18).

e. FindP(8<x<18).

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g. What is the probability that xis in the intervalμ±2σ?

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