Prove Whether A Person's Skill Level By A Simple Computation

By A Simple Computation Prove Whether A Persons Skill Level Is Indepe

Evaluate whether a person's skill level is independent of their voting inclination using the provided chart. The chart details the distribution of skilled and unskilled individuals in relation to their voting attitudes—namely, against, no opinion, and total for each group.

Calculate the probability that a standard normal variable is less than or equal to 0.2.

Given a normal distribution with a mean (μ) of 3 and standard deviation (σ) of 0.5, determine the probability that a random variable X assumes a value less than or equal to 2.3.

Suppose friends decide to "go for a drink" after an exam. Find the probability that the waitress refuses service to exactly 2 minors, given that she randomly checks IDs of 5 students out of 9, where 4 are minors.

Analyze a dataset where a random sample of 100 recorded deaths in the United States shows an average life span of 71.8 years with a standard deviation of 8.9 years. Apply a significance level of 0.05 to assess whether this suggests the current mean life span exceeds 70 years.

Calculate the 90% confidence interval for the average life span based on the data provided.

Compare two soft-drink machines by analyzing independent samples of 45 and 55 bottles respectively. The sample means are 0.84 oz and 0.91 oz, with standard deviations of 0.046 oz and 0.044 oz. Test at a 0.05 significance level whether the mean content from machine 1 is statistically smaller than that from machine 2.

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The question of whether an individual's skill level is independent of their voting inclination can be examined through statistical independence testing. Using categorical data, such as the chart with counts of skilled/unskilled persons and their voting attitudes, we can perform a chi-square test of independence to assess whether the variables are associated or independent. This involves calculating expected frequencies under the assumption of independence and comparing them to observed counts to evaluate the null hypothesis that skill level and voting inclination are independent.

Furthermore, probability calculations related to the standard normal distribution help understand variability and standardize scores. For instance, finding P(Z ≤ 0.2) involves calculating the cumulative distribution function (CDF) value for a standard normal variable Z. Since Z follows a standard normal distribution with mean 0 and variance 1, this probability can be obtained via standard normal tables or computational tools, which yield approximately 0.5793, indicating the probability that Z is less than or equal to 0.2.

Similarly, for a normal distribution with μ=3 and σ=0.5, the probability P(X ≤ 2.3) is computed by standardizing the variable to obtain the Z-score: Z = (2.3 - 3)/0.5 = -1.4. Using the standard normal table, we find P(Z ≤ -1.4) ≈ 0.0808, indicating an approximately 8.08% chance that X is less than or equal to 2.3.

Regarding the scenario where friends go for drinks, the probability that exactly 2 minors are refused service (assuming the waitress checks 5 students randomly from among 9, with 4 minors), can be modeled using the hypergeometric distribution. The hypergeometric probability is calculated as:

P(X=2) = [(C(4,2) C(5,3))] / C(9,5) = [(6 10)] / 126 = 60/126 ≈ 0.476. This suggests that there is approximately a 47.6% chance exactly 2 minors will be refused service under these conditions.

The analysis of the U.S. mortality data involves conducting a one-sample t-test to determine if the current mean lifespan exceeds 70 years. With a sample mean of 71.8 years, a standard deviation of 8.9 years, and a sample size of 100, the test statistic t is computed as:

t = (71.8 - 70) / (8.9 / √100) = 1.8 / (8.9/10) = 1.8 / 0.89 ≈ 2.022. Comparing this to the critical t-value for 99 degrees of freedom at alpha=0.05 (which is approximately 1.984), we see that 2.022 > 1.984, leading us to reject the null hypothesis and conclude there is evidence that the mean lifespan has increased beyond 70 years.

The 90% confidence interval for the mean lifespan is calculated using the formula:

CI = x̄ ± t(s / √n), where t is the critical value for 90% confidence with df=99 (~1.66). Plugging in the values:

CI = 71.8 ± 1.66 (8.9 / 10) = 71.8 ± 1.66 0.89 ≈ 71.8 ± 1.48,

which results in an interval approximately from 70.32 to 73.28 years.

Comparing the means of soft drinks from two machines involves a two-sample t-test for unequal variances. The test statistic is computed as:

t = (X̄1 - X̄2) / sqrt( (s1^2 / n1) + (s2^2 / n2) ),

which with values becomes:

t = (0.84 - 0.91) / sqrt( (0.046^2 / 45) + (0.044^2 / 55) ) ≈ -0.07 / sqrt( (0.002116 / 45) + (0.001936 / 55) ) ≈ -0.07 / sqrt( 0.000047 + 0.000035 ) ≈ -0.07 / sqrt(0.000082) ≈ -0.07 / 0.00906 ≈ -7.72.

Since the t-value is large in magnitude and negative, and assuming the degrees of freedom are sufficient, we reject the null hypothesis at 0.05 significance level, establishing that the mean content of machine 1 is statistically smaller than that of machine 2.

These statistical analyses demonstrate the application of probability distributions, hypothesis testing, and interval estimation in diverse real-world contexts. They show how statistical methods can inform decision-making, from evaluating independence between social factors to comparing product qualities and interpreting mortality data.

References

  • Bluman, A. G. (2018). Elementary Statistics: A Step By Step Approach. McGraw-Hill Education.
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