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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

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’s expected profit per metal part produced.

If a population data set is normally distributed, what isthe proportion of measurements you would expect to fallwithin the following intervals?

a.μ±σb.μ±2σc.μ±3σ

Assume that xis a random variable best described by a uniform distribution with c=10andd=90.

a. Findf(x).

b. Find the mean and standard deviation of x.

c. Graph the probability distribution for xand locate its mean and theintervalon the graph.

d. FindP(x≤60).

e. FindP(x≥90).

f. FindP(x≤80).

g. FindP(μ-σ≤x≤μ+σ).

h. FindP(x>75).

The random variable x has a normal distribution with μ=1000 and σ=10.

a. Find the probability that x assumes a value more than 2 standard deviations from its mean. More than 3 standard deviations from .μ

b. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviations of μ.

c. Find the value of x that represents the 80th percentile of this distribution. The 10th percentile.

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).

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