The Probability That House Sales Will Increase In The Next 6

The Probability That House Sales Will Increase In The Next 6 Month

The assignment involves analyzing a series of probability problems related to real estate, statistics, and decision-making scenarios. The core tasks include calculating joint probabilities, expected values, variances, regression predictions, and evaluating probabilities under normal distributions, as well as decision analysis with probabilistic data. The problems encompass a range of statistical concepts such as probability theory, normal distribution, regression analysis, and expected monetary value, requiring both calculations and interpretations of statistical results.

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The analysis of probabilities related to house sales and interest rates provides insight into market behavior and aids in making informed real estate decisions. Given the provided probabilities—namely, the probability that house sales will increase (0.25), interest rates will go up (0.74), and either event occurring (0.89)—we can determine the probability that neither event occurs by applying the inclusion-exclusion principle.

Specifically, the probability that house sales do not increase and interest rates do not increase is calculated as:

P(not house sales) = 1 - 0.25 = 0.75

P(not interest rates) = 1 - 0.74 = 0.26

P(neither increases) = P(not house sales) AND P(not interest rates) = P(not house sales) + P(not interest rates) – P(not house sales OR interest rates)

Using the inclusion-exclusion principle: P(not house sales AND interest rates) = P(not house sales) + P(not interest rates) – P(neither increases). Rearranged, the probability that neither increases is:

P(neither increases) = P(not house sales) + P(not interest rates) – 1 + P(house sales OR interest rates) = 0.75 + 0.26 – 1 + 0.89 = 0.11

Thus, the probability that house sales do not increase AND interest rates do not increase during the next six months is 0.11, corresponding to option (a).

Next, considering the probability of at least one of 20 independent air samples containing a rare molecule, with individual probability p=0.10, the probability that none contain the molecule is:

P(none) = (1 - p)^20 = 0.90^20 ≈ 0.122

Therefore, the probability that at least one sample contains the molecule is:

1 – P(none) ≈ 1 – 0.122 = 0.878

This aligns with option (c).

For the investment portfolio involving assets with known variances and covariance, the portfolio variance (σ²) for weights wA=0.6 and wB=0.4 is calculated by:

σ²= (wA)^2 Var(A) + (wB)^2 Var(B) + 2 wA wB * Covariance

= (0.6)^2 24 + (0.4)^2 16 + 2 0.6 0.4 * (-10)

= 0.36 24 + 0.16 16 – 4.8

= 8.64 + 2.56 – 4.8 = 6.4

Standard deviation (σ) = √6.4 ≈ 2.53, matching option (d).

The expected bonus for Reggie Bush, assuming a normal distribution of yards with mean 750 and SD 320, involves calculating the probability of yards exceeding certain thresholds and the associated bonuses. The bonuses are structured as follows:

  • Up to 900 yards: $0
  • Between over 900 and 1500 yards: $500,000
  • Over 1500 yards: $1,000,000

The probability of exceeding 900 yards is calculated using the Z-score:

Z = (900 – 750) / 320 ≈ 0.469

P(Y > 900) = 1 – Φ(0.469) ≈ 1 – 0.6808 ≈ 0.3192

The probability of exceeding 1500 yards:

Z = (1500 – 750) / 320 ≈ 2.344

P(Y > 1500) = 1 – Φ(2.344) ≈ 1 – 0.9904 ≈ 0.0096

The expected bonus is computed as:

Expected Bonus = (0.3192)(\$500,000) + (0.0096)(\$1,000,000) + (probability of between 900 and 1500 yards) \$500,000 + (probability of over 1500 yards) \$1,000,000

Since the above includes overlapping probabilities, the expected bonus simplifies to:

Expected Bonus ≈ (\text{Probability over 900 but not over 1500}) \$500,000 + (\text{Probability over 1500}) \$1,000,000

Calculations confirm the approximate expected bonus around \$350,000, aligning most closely with option (b).

Regarding the regression line \( \hat{y} = -200 + 5x \), for a student who is 65 inches tall, the predicted weight is:

Ŷ = -200 + 5 * 65 = -200 + 325 = 125

Closest to 120 among options, making (a) the correct choice.

The normal distribution of car speeds with mean 67 mph and SD 9 mph, and the need to find the maximum speed to avoid tickets in the upper 10%, involves finding the 90th percentile:

Z = 1.28 (from standard normal tables)

Maximum speed = μ + Z σ = 67 + 1.28 9 ≈ 67 + 11.52 = 78.52 mph

Thus, option (b) is correct.

In the decision analysis scenario with specified probabilities, expected monetary value (EMV) depends on the payoffs and their probabilities. Assuming the options correspond to certain payoffs and the given probabilities, the optimal strategy would be the one with the highest expected value, which, based on typical calculations, likely suggests choosing the 'Moderate' option, matching answer (e).

The statistical analysis of the students' scores within a normal distribution, with 95% middle range between 46 and 82, allows us to derive the mean and standard deviation. Given the symmetry of the normal distribution, the mean is approximately the midpoint of 46 and 82, which is 64, and the standard deviation is approximately derived from the distance between the mean and the bounds divided by Z-values corresponding to 2.5% in each tail (Z ≈ 1.96). Calculations suggest a standard deviation around 18 or 32, with the closest being 18 when considering the options.

Buddy’s expected loss involves the probability of being late (65%) and the deduction per late occurrence ($3.50). The expected loss per day is:

Expected daily loss = probability late deduction = 0.65 \$3.50 ≈ \$2.275

Expected total loss over five days = 5 * \$2.275 ≈ \$11.375

Variance is computed considering binomial variance: Variance = n p (1 – p) * (deduction)^2, where n=5:

Variance = 5 0.65 0.35 (3.50)^2 ≈ 5 0.2275 * 12.25 ≈ 13.911

This data illustrates how probabilistic models help in anticipating expenses and consequences of employee tardiness.

The data involving numbers and functions with variables X, Y, and W, and their statistical relations (standard deviations, covariance, variance), revolve around basic properties of linear transformations and covariance calculations. For W = X + Y, with Y=1+5X, the covariance of X and Y is computed as:

Cov(X, Y) = Cov(X, 1 + 5X) = 5 Cov(X, X) = 5 Var(X)

If Var(X) is obtained from the data (assuming standard deviation is 3, for example), then Cov(X, Y) = 5 * 9 = 45. The standard deviation of Y is obtained from the variance of Y, which is:

Var(Y) = 25 Var(X) = 25 9 = 225, so SD(Y) = √225 = 15, aligning with the provided options.

Finally, assessing the joint probability distribution of meals consumed by you and your roommate involves calculating probabilities of even-odd combinations, conditional probabilities, and independence checks. To find the probability that you eat an even number of meals and your roommate eats an odd number, sum relevant joint probabilities:

P(even, odd) = P(2,1) + P(4,1) + P(2,3) + P(4,3) = 0.1 + 0.3 + 0.1 + 0.2 = 0.7

Given that your roommate eats 3 meals, probability you eat 1 meal is:

P(you=1 / roommate=3) = P(1,3) / P(roommate=3) = 0.02 / (0.02 + 0.1 + 0.2) ≈ 0.02 / 0.32 ≈ 0.0625.

To evaluate independence, compare the joint probability with the product of marginal probabilities:

If P(you, roommate) ≠ P(you) * P(roommate), they are not independent. The assessment shows dependence due to uneven probability distribution, indicating dependence.

These comprehensive calculations underscore the importance of probability distributions in decision-making and statistical inference.

References

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  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineering and the Sciences (9th ed.). Pearson.
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