Suppose You Invest A Fixed Sum Of Money In Each Of Five Inve
Suppose You Invest A Fixed Sum Of Money In Each Of Five Internet B
Suppose you invest a fixed sum of money in each of five internet business ventures. Assume you know that 70% of each venture is successful, the outcomes of the ventures are independent of one another, and the probability distribution for the number, x, of the successful ventures out of five is given by the binomial distribution. You are asked to calculate the mean, variance, and standard deviation of this distribution. Additionally, for a binomial random variable with n=6 and p=0.37, you are required to compute specific probabilities and summary statistics. Finally, for GMAT scores distributed normally with a mean of 500 and standard deviation of 100, you need to determine the probabilities that scores fall within certain ranges.
Paper For Above instruction
The analysis of binomial distributions is fundamental in understanding the probabilistic outcomes of independent trials, such as investments across multiple ventures or evaluation scores. In this context, the problem revolves around the binomial random variable representing the number of successful ventures out of five attempts, each with a success probability of 0.7. Subsequently, the problem extends to calculations involving the binomial distribution with different parameters, as well as the application of normal distribution principles to standardized test scores.
Part 1: Binomial Distribution with Five Ventures
The problem specifies that five independent internet ventures are undertaken, each with a success probability of p = 0.7. The random variable X, denoting the number of successful ventures out of these five, follows a binomial distribution B(n=5, p=0.7). The probability mass function (PMF) for this distribution is given by:
\[
P(X = x) = \binom{5}{x} (0.7)^x (0.3)^{5 - x}
\]
for x = 0, 1, 2, 3, 4, 5.
(a) Calculation of the Mean (μ):
The mean of a binomial distribution is calculated by:
\[
\mu = n \times p = 5 \times 0.7 = 3.5
\]
Thus, the expected number of successful ventures is 3.5.
(b) Calculation of the Variance:
The variance of a binomial distribution is:
\[
\sigma^2 = n \times p \times (1 - p) = 5 \times 0.7 \times 0.3 = 0.75
\]
(c) Calculation of the Standard Deviation:
The standard deviation is the square root of the variance:
\[
\sigma = \sqrt{0.75} \approx 0.866
\]
Part 2: Binomial Probabilities with n=6 and p=0.37
For a binomial random variable X with parameters n=6 and p=0.37, various probabilities are requested.
(a) \( P(X=2) \):
Using the binomial PMF:
\[
P(X=2) = \binom{6}{2} (0.37)^2 (0.63)^4
\]
Calculating:
\[
\binom{6}{2} = 15
\]
\[
(0.37)^2 \approx 0.1369
\]
\[
(0.63)^4 \approx 0.1575
\]
Thus,
\[
P(X=2) \approx 15 \times 0.1369 \times 0.1575 \approx 15 \times 0.0216 \approx 0.324
\]
(b) \( P(X > 4) \):
This is the sum of probabilities for X=5 and X=6:
\[
P(X>4) = P(X=5) + P(X=6)
\]
Calculations:
\[
P(X=5) = \binom{6}{5} (0.37)^5 (0.63)^1 = 6 \times (0.37)^5 \times 0.63
\]
List of computations:
\[
(0.37)^5 \approx 0.0052
\]
\[
(0.63)^1 = 0.63
\]
Thus,
\[
P(X=5) \approx 6 \times 0.0052 \times 0.63 \approx 6 \times 0.0033 \approx 0.0198
\]
Similarly,
\[
P(X=6) = \binom{6}{6} (0.37)^6 (0.63)^0 = 1 \times (0.37)^6 \times 1
\]
\[
(0.37)^6 \approx 0.0019
\]
Therefore,
\[
P(X > 4) \approx 0.0198 + 0.0019 = 0.0217
\]
(c) \( P(X \leq 3) \):
This is the cumulative probability for X=0,1,2,3. Using binomial table or calculator simplifies this process. Based on binomial probabilities, the sum yields approximately 0.820.
Part 3: Summary Statistics of Binomial Variable
The mean (μ) and variance (σ²) for the binomial distribution with n=6, p=0.37 are for example calculated as:
\[
\mu = n \times p = 6 \times 0.37 = 2.22
\]
\[
\sigma^2 = n \times p \times (1 - p) = 6 \times 0.37 \times 0.63 \approx 1.40
\]
and the standard deviation:
\[
\sigma = \sqrt{1.40} \approx 1.183
\]
Part 4: Normal Distribution of GMAT Scores
The scores are normally distributed with a mean (μ) of 500 and a standard deviation (σ) of 100. Probabilities of scores within certain ranges are calculated using the standard normal distribution Z = (X - μ)/σ.
(a) Proportion between 500 and 550:
Calculate Z-scores:
\[
Z_{500} = (500 - 500)/100 = 0
\]
\[
Z_{550} = (550 - 500)/100 = 0.5
\]
Using standard normal distribution tables, P(Z between 0 and 0.5) ≈ 0.1915. Since Z=0 corresponds to the mean, the proportion of scores between 500 and 550 is approximately 19.15%.
(b) Proportion between 450 and 600:
Z-scores:
\[
Z_{450} = (450 - 500)/100 = -0.5
\]
\[
Z_{600} = (600 - 500)/100 = 1.0
\]
From tables, P(Z between -0.5 and 1.0) ≈ 0.6915, indicating about 69.15% of scores fall within this range.
(c) Proportion between 550 and 750:
Z-scores:
\[
Z_{550} = 0.5
\]
\[
Z_{750} = 2.5
\]
From the standard normal tables, P(Z between 0.5 and 2.5) ≈ 0.4938, meaning roughly 49.38% of scores are in this interval.
Conclusion
These calculations demonstrate the application of binomial and normal distribution principles to practical scenarios, including investment success probabilities and standardized test scores. Understanding these distributions enables better decision-making and risk assessment in fields such as finance, education, and research.
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