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navik [9.2K]
3 years ago
10

. Consumer Awareness In 1986, the average cost of a new midsize four-door sedan was $9000. In 1991, the average cost was $12,000

. Determine an exponential model for predicting the cost of a new midsize car. Use the model to predict the cost of a comparable new car in the year 2000.
Mathematics
1 answer:
eduard3 years ago
8 0

<u>Answer-</u>

\boxed{\boxed{y=9000(1.0592)^x}}

<em>And cost of a car in the year 2000 will be </em><em>$20134.18</em>

<u>Solution-</u>

Let's assume,

x = number of years after 1986

y = average cost of sedan in dollar

The exponential model that will model the scenario will be in the form of,

y=ab^x

where a and b are constants

As given that, in 1986 average cost of sedan was $9000 and in 1991 average cost of sedan was $12000

So, the points (0, 9000) and (5, 12000) will satisfy or lie on the exponential curve.

Putting (0, 9000) in the equation,

\Rightarrow 9000=ab^0

\Rightarrow a\times 1=9000

\Rightarrow a=9000

Now, the equation becomes y=9000b^x

Putting (5, 12000) in this equation,

\Rightarrow 12000=9000b^5

\Rightarrow b^5=\dfrac{12000}{9000}

\Rightarrow b=\sqrt[5]{\dfrac{4}{3}}

\Rightarrow b=1.0592

Putting the values,

y=9000(1.0592)^x

As we have to calculate the cost of sedan in 2000, so putting x=14(as 2000-1986=14),

y=9000(1.0592)^{14}=20134.18

Therefore, cost of a car in the year 2000 will be $20134.18

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Answer:

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b) probability the player loses $1000 or more is 0.1788

c)  probability the player wins is 0.3557

d) probability of going broke is 0.0594

Step-by-step explanation:

Given:

Since there are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 * 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

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Solution:

Expected payoff is basically the expected mean

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss =  μ

                        = ∑ x P(x)

                        = 50 * P(win) - 50 * P(lose)

                        = 50 * P(win) + (-50) * (1 - P(win))

                         = 50 * 0.49 - 50 * (1 - 0.49)

                        = 24.5 - 50 ( 0.51 )

                        = 24.5 - 25.5

                        = -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 * 1 = $ 240

b)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 110.5 - 117.6 / √117.6(1-0.49)

  = −7.1/√117.6(0.51)

  = −7.1/√59.976

  = −7.1/7.744417

  =−0.916789

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110

Using normal probability table:

P(X≤110) = P(X<110.5)

             = P(Z<-0.916)

             = 0.1788

c)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 120.5 - 117.6 / √117.6(1-0.49)

  = 2.9/√117.6(0.51)

  = 2.9/√59.976

  = 2.9/7.744417

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Here the player wins when he wins at least 120 of 240 hands

Using normal probability table:

P(X>120) = P(X>120.5)

              = P(Z>0.3744)  

             =  1 - P(Z<0.3744)

             = 1 - 0.6443

             = 0.3557

d)

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Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

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Compute z-score:

z = x - np / √(np(1-p))

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  = -12.1/√59.976

  = -12.1/7.744417

  =−1.562416

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

             = P(Z<-1.562)

             = 0.0594

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