In China, 4-Year-Olds Average 3 Hours A Day Unsupervised
1in China 4 Year Olds Average 3 Hours A Day Unsupervised Most Of Th
In China, 4-year-olds average 3 hours a day unsupervised. Most of these children live in rural areas, which are considered safe. The amount of time spent alone is normally distributed with a standard deviation of 1.5 hours. We are asked to analyze specific cases of children and their unsupervised time, and to compute probabilities based on the normal distribution.
Firstly, a random Chinese 4-year-old is observed to spend 4.25 hours a day unsupervised. We need to determine how many standard deviations this value is from the mean. Secondly, we will calculate the probability that a child spends more than 4.25 hours alone daily. Conversely, we will examine a scenario where another child spends only 1 hour unsupervised per day and determine how many standard deviations this is from the mean, along with the corresponding probability that a child spends less than 1 hour alone.
Additionally, the problem extends to a different context involving age in China, where the mean age is 16 with a standard deviation of 4.5. We are to find the probability that a randomly selected individual is aged between 21 and 25. Finally, a scenario involving salary raises is presented; the mean raise is $2000 with a standard deviation of $400, assumed normally distributed. We are tasked with identifying the minimum raise for the top 5.05% most productive employees and the maximum raise for the bottom 10.2% least productive employees.
Paper For Above instruction
Analysis of Child Unsupervised Time in China
The normal distribution provides a powerful framework for analyzing data related to human behaviors and characteristics, especially when such data tend to cluster around a central mean with symmetrical spread. In the case of 4-year-old children in China, understanding the typical amount of unsupervised time they experience offers insights into childcare practices, safety perceptions, and social norms. This analysis explores how specific data points relate to the overall distribution, probabilities of certain events, and broader biometric and sociological implications.
Part A: Deviation of 4.25 Hours from the Mean
Given the mean (μ) of 3 hours and a standard deviation (σ) of 1.5 hours, the z-score for a child spending 4.25 hours alone per day is calculated as:
z = (X - μ) / σ = (4.25 - 3) / 1.5 = 1.25 / 1.5 ≈ 0.83
This z-score of approximately 0.83 indicates that the child's unsupervised time is 0.83 standard deviations above the mean, placing it within the typical upper range of children's behaviors but still well within the expected variation of the normal distribution.
Part B: Probability of Spending More Than 4.25 Hours Unsupervised
Using the standard normal distribution table (Z-table), the probability that a z-score exceeds 0.83 is:
P(Z > 0.83) ≈ 1 - P(Z
Therefore, there is approximately a 20.33% chance that a randomly selected rural Chinese 4-year-old spends more than 4.25 hours unsupervised per day.
Part C: Deviation of 1 Hour from the Mean
Calculating the z-score for a child spending only 1 hour unsupervised:
z = (1 - 3) / 1.5 = -2 / 1.5 ≈ -1.33
This indicates that the child's unsupervised time is 1.33 standard deviations below the mean, suggesting they spend significantly less time alone than average but still falling within the expected variation.
Part D: Probability of Spending Less Than 1 Hour Unsupervised
Using the Z-table, the probability that z is less than -1.33:
P(Z
Thus, approximately 9.18% of rural Chinese 4-year-olds spend less than 1 hour unsupervised daily.
Analysis of Age Distribution in China
In another context, suppose the population’s age in China is normally distributed with a mean of 16 years and a standard deviation of 4.5 years. To find the probability that a randomly selected individual is between 21 and 25 years of age, we calculate the respective z-scores:
z for 21 years: (21 - 16) / 4.5 ≈ 1.11
z for 25 years: (25 - 16) / 4.5 ≈ 2.00
Consulting the Z-table, the probabilities corresponding to these z-scores are:
P(Z
P(Z
The probability that an individual’s age falls between 21 and 25 is the difference between these two probabilities:
0.9772 - 0.8665 = 0.1107
Therefore, there is approximately an 11.07% chance that a randomly selected individual in China is aged between 21 and 25 years.
Salary Raise Distribution Analysis
The organization plans to distribute salary raises with a normal distribution where the mean is $2000, and the standard deviation is $400. To identify the cutoffs for the top 5.05% and bottom 10.2%, we determine the respective z-scores using standard normal distribution tables or software.
For the upper tail at 5.05%, the corresponding z-score is approximately 1.66, since P(Z > 1.66) ≈ 0.05. The minimum raise for the most productive 5.05% is then:
Raise = μ + z σ = 2000 + 1.66 400 ≈ 2000 + 664 = $2664
For the lower tail at 10.2%, the corresponding z-score is approximately -1.23, since P(Z
Raise = μ + z σ = 2000 - 1.23 400 ≈ 2000 - 492 = $1508
These values help management identify thresholds for reward distribution according to productivity based on normal distribution principles.
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