Let Z Be A Standard Normal Random Variable: Calculate The Fo ✓ Solved

Let Z Be Standard Normal Random Variable Calculate The Following

Calculate the following probabilities related to a standard normal random variable Z, rounding your responses to at least three decimal places:

  • a. P(Z > -1.97)
  • b. P(Z ≤ -1.71)
  • c. P(1.01

Determine the value of c such that:

  • a. P(-c ≤ Z ≤ c) = 0.9700; round your answer to two decimal places.
  • b. P(c ≤ Z ≤ -1.22) = 0.0745; round your answer to two decimal places.

Suppose the quarterly sales levels among healthcare information systems companies are approximately normally distributed with a mean of 12 million dollars and a standard deviation of 1.3 million dollars. A quarter is considered a "failure" if its sales are in the bottom 10%. Determine the sales level (in millions of dollars) that separates failing quarters from non-failing ones, rounding your answer to one decimal place.

A psychologist has designed a questionnaire measuring individuals' aggressiveness. The scores are normally distributed with an unknown mean and a standard deviation of 80. If exactly 5% of scores exceed 800, find the mean score, rounding to at least one decimal place.

An aptitude test measuring leadership abilities has scores normally distributed with a mean of 560 and a standard deviation of 125. The potential leaders are those who score above 700. Calculate the proportion of the population that qualify as potential leaders, rounding your answer to at least four decimal places.

A multiple-choice exam with 80 questions, each with options a, b, c, d, and e, involves guessing answers randomly. Approximate the probability that a student guesses at least 13 answers correctly using the normal approximation to the binomial distribution with a continuity correction. Round your answer to three decimal places.

A worldwide organization claims their members have a mean IQ score of 118 with a standard deviation of 16. A random sample of 40 members yields a sample mean IQ of 114.2. Determine the probability of observing a sample mean of 114.2 or less, assuming the claim is true. Round your answer to three decimal places.

Regarding stock market data, 600 stocks that went up yesterday have a 60% chance of going up again today. Find:

  • a. The expected proportion p of stocks that will go up today.
  • b. The standard deviation of p.
  • c. The probability that more than 58% of the stocks will go up today (p > 0.58), approximated using the normal distribution, rounded to four decimal places.

Paper For Above Instructions

The following comprehensive analysis addresses the variety of statistical problems based on the standard normal distribution, probability calculations, and applications of normal approximation to binomial distributions. These problems are foundational in understanding probability theory and statistical inference in real-world contexts.

1. Calculating Probabilities for a Standard Normal Variable

Using the standard normal distribution (Z), the probabilities are derived from Z-tables or statistical software. For example, P(Z > -1.97) can be found as 1 - P(Z ≤ -1.97), but since the normal distribution is symmetric, P(Z > -1.97) = P(Z

2. Determining the Value of c for Symmetrical Confidence Interval

To find c such that P(-c ≤ Z ≤ c) = 0.97, we identify the Z-score corresponding to the middle 97%, leaving 1.5% in each tail. From Z-tables, these are approximately ±2.17. For the second problem, where P(c ≤ Z ≤ -1.22) = 0.0745, the Z-score for 0.0745 in the upper tail corresponds to approximately ±1.22; adjusting as required yields c ≈ 1.50.

3. Sales Level Representing Bottom 10%

To find the sales level indicating the 10th percentile, we find the Z-value corresponding to the 10% lower tail, which is approximately -1.28. Applying the z-score formula: x = μ + zσ, yields 12 + (-1.28)(1.3) ≈ 12 - 1.664 = 10.3 million dollars.

4. Estimating the Mean of Aggressiveness Scores

Given that 5% of scores exceed 800, the Z-score for this upper 5% tail is approximately 1.645. Using x = μ + zσ, we rearrange to find μ = x - zσ = 800 - 1.645(80) ≈ 800 - 131.6 = 668.4.

5. Proportion of Potential Leaders

With mean 560 and SD 125, those scoring above 700 are considered leaders. The Z-score is (700 - 560)/125 = 1.12. Using the standard normal table, the area to the right of Z=1.12 is approximately 0.1314, indicating about 13.14% of the population qualify as potential leaders.

6. Normal Approximation to Binomial Distribution

For 80 questions with probability p=0.2 of correct guesses, the mean number of correct answers is np=16, and variance np(1-p)=12.8. To find P(X ≥ 13), we apply a continuity correction and convert to Z: Z = (12.5 - 16)/√12.8 ≈ -1.06. Therefore, P(X ≥ 13) ≈ P(Z ≥ -1.06) ≈ 0.8554.

7. Probability of Sample Mean IQ ≤ 114.2

The standard error (SE) = σ/√n = 16/√40 ≈ 2.53. Calculating Z = (114.2 - 118)/2.53 ≈ -1.48. The corresponding probability, P(Z ≤ -1.48), is approximately 0.0694, so there's about a 6.94% chance the sample mean IQ is 114.2 or less.

8. Stock Market Data – Proportion and Variance

The expected proportion p̄ equals the population proportion, thus p̄=0.6. The standard deviation of p̄ is √(p(1-p)/n) = √(0.6×0.4/600) ≈ 0.0201. To find P(p > 0.58), compute Z = (0.58 - 0.6)/0.0201 ≈ -0.995. Using the normal distribution, P(Z > -0.995) ≈ 0.84, indicating an approximate 84% chance that more than 58% of these stocks will go up again today.

Conclusion

The problems explored demonstrate the power of the normal distribution in statistical inference, from probability calculations to hypothesis testing and confidence intervals. These techniques are essential in various fields, including finance, psychology, healthcare, and organizational behavior, providing vital insights based on data assumptions and distribution properties.

References

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