A coin is weighted so that the probability of obtaining a head in a single toss is 0.3. If the coin is tossed 35 times, what is the probability of obtaining between 9 and 14 heads, exclusive. a. 0.510 b. 0.627 c. 0.701 d. 0.287 e. 0.824 9. If 14% of men are bald, what is the probability that more than 100 in a random sample of 850 men are bald?a. 0.9778
b. 0.9664
c. 0.0697
d. 0.0027
e. 0.0378The Colorado Mining and Mineral Company has 1000 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-yr period is 0.08. What is the probability that more than 70 workers will meet with an accident during the 1-yr period?
a. 0.8665
b. 0.9089
c. 0.3585
d. 0.3827
e. 0.5753

Respuesta :

Answer:

(1) The correct answer is option (a) 0.510

(2) The correct answer is option (b) = 0.9664

(3) The correct answer is option (a)  0.8665

Step-by-step explanation:

1. Find attached of question 1

2.

n= 850

p= 0.1400  

here mean of distribution(μ) =n*p

                                               =850* 0.1400

                                                = 119  

standard deviation σ = √(np(1-p))

                                    = √(850*0.1400(1-0.1400))

                                    = √102.34

                                       =10.1163  

for normal distribution z score =(X-μ)/σx    

therefore from normal approximation of binomial distribution and continuity correction:

probability = P(X>100.5)

                   = P(Z>-1.83)

                   = 1-P(Z<-1.83)

                   = 1-0.0336

                   = 0.9664

3.

n = 1000

p = 0.08

mean of distribution(μ) =np

                                       = 1000*0.08

                                        = 80

and standard deviation σ=√(np(1-p))

                                    = √(1000*0.08(1-0.08))

                                    = √736

                                    = 8.5790

probability = P(X>70.5)

                   = P(Z>-1.11)

                    = 1-P(Z<-1.11)

                    = 1-0.1335

                    = 0.8665

Ver imagen gbenewaeternity

The three questions are illustrations of normal approximation of binomial distribution and continuity correction

(a) Coins

The given parameters are:

[tex]\mathbf{n =35}[/tex]

[tex]\mathbf{p =0.3}[/tex]

Start by calculating the mean and the standard deviation

[tex]\mathbf{\bar x = np = 35 \times 0.3 = 10.5}[/tex]

[tex]\mathbf{\sigma = \sqrt{\bar x(1 - p)} = \sqrt{10.5 \times (1 - 0.3 )}= 2.71}[/tex]

The probability is then represented as:

[tex]\mathbf{P(9 < x < 14) = P(8.5 < x < 14.5)}[/tex]

Calculate the z-scores for x = 8.5, and 14.5

[tex]\mathbf{z = \frac{x - \bar x}{\sigma}}[/tex]

So, we have:

[tex]\mathbf{z = \frac{8.5 - 10.5}{2.71} = -0.3688}[/tex]

[tex]\mathbf{z = \frac{14.5 - 10.5}{2.71} = 1.1066}[/tex]

So, we have:

[tex]\mathbf{P(9 < x < 14) = P(-0.3688<z<1.1066)}[/tex]

Rewrite as:

[tex]\mathbf{P(9 < x < 14) = P(z<1.1066) - P(z<-0.3688)}[/tex]

Using z-scores of probabilities, we have:

[tex]\mathbf{P(9 < x < 14) = 0.86577- 0.3561}[/tex]

[tex]\mathbf{P(9 < x < 14) = 0.50967}[/tex]

Approximate

[tex]\mathbf{P(9 < x < 14) = 0.510}[/tex]

Hence, the probability of obtaining between 9 and 14 heads (exclusive) is (a) 0.510

(b) Bald men

The given parameters are:

[tex]\mathbf{n =850}[/tex]

[tex]\mathbf{p =14\%}[/tex]

Start by calculating the mean and the standard deviation

[tex]\mathbf{\bar x = np = 850 \times 14\% = 119}[/tex]

[tex]\mathbf{\sigma = \sqrt{\bar x(1 - p)} = \sqrt{119 \times (1 - 14\% )}= 10.12}[/tex]

The probability is then represented as:

[tex]\mathbf{P(x > 100) = P(x > 100.5)}[/tex]

Calculate the z-scores for x = 100.5

[tex]\mathbf{z = \frac{x - \bar x}{\sigma}}[/tex]

So, we have:

[tex]\mathbf{z = \frac{100.5 - 119}{10.12} = -1.83}[/tex]

So, we have:

[tex]\mathbf{P(x > 100) = P(z > -1.83)}[/tex]

Using z-scores of probabilities, we have:

[tex]\mathbf{P(x > 100) = 0.96638}[/tex]

Approximate

[tex]\mathbf{P(x > 100) = 0.9664}[/tex]

Hence, the probability that more than 100 in 850 are bald is (b) 0.9664

(c) Accidents

The given parameters are:

[tex]\mathbf{n =1000}[/tex]

[tex]\mathbf{p =0.08}[/tex]

Start by calculating the mean and the standard deviation

[tex]\mathbf{\bar x = np = 1000 \times 0.08 = 80}[/tex]

[tex]\mathbf{\sigma = \sqrt{\bar x(1 - p)} = \sqrt{80 \times (1 - 0.08)}= 8.58}[/tex]

The probability is then represented as:

[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex]

Calculate the z-scores for x = 70.5

[tex]\mathbf{z = \frac{x - \bar x}{\sigma}}[/tex]

So, we have:

[tex]\mathbf{z = \frac{70.5 - 80}{8.58} = -1.11}[/tex]

So, we have:

[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]

Using z-scores of probabilities, we have:

[tex]\mathbf{P(x > 70) =0.8665}[/tex]

Hence, the probability that more than 70 workers will be involved in an accident is (a) 0.8665

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