Problem 29 - Calculus Concepts

Question

Learning Time The manufacturer of a new board game believes that the time it takes a child between the ages of 8 and 10 to learn the rules of this game has the probability density function \(g(t)=\left\\{\begin{array}{ll}\frac{3}{32}\left(4 t-t^{2}\right) & \text { when } 0 \leq t \leq 4 \\ 0 & \text { elsewhere }\end{array}\right.\) where \(t\) is time measured in minutes. a. Calculate the mean time it takes a child age 8 to 10 to learn the rules of this game. b. Calculate the standard deviation of the learning times. c. Calculate \(P(t \leq 3)\). Interpret this result.

Step-by-Step Solution

Answer

The mean is 2 minutes, the standard deviation is 0.816, and \(P(t \leq 3) = 0.84375\). This means there's an 84.375% chance a child learns the game in 3 minutes or less.

Step 1: Understand the Problem

We are given a probability density function (PDF) \( g(t) \) for the time \( t \) in minutes that a child takes to learn the rules of a board game. This PDF is valid for \( 0 \leq t \leq 4 \) and zero elsewhere. We need to find (a) the mean, (b) the standard deviation of \( t \), and (c) \( P(t \leq 3) \).

Step 2: Calculate the Mean

The mean of a continuous random variable with a given density function \( g(t) \) is calculated using the formula \( \mu = \int_{-\infty}^{\infty} t \, g(t) \, dt \). For this problem, the integration bounds are from 0 to 4, as the PDF is zero elsewhere. Therefore, compute the integral:\[\mu = \int_{0}^{4} t \left( \frac{3}{32}(4t - t^2) \right) \, dt\]

Step 3: Solve the Mean Integral

Evaluate the integral for the mean:\[\mu = \frac{3}{32} \int_{0}^{4} (4t^2 - t^3) \, dt = \frac{3}{32} \left[ \frac{4}{3}t^3 - \frac{1}{4}t^4 \right]_{0}^{4}\]Plug in the limits to find the result of this computation.

Step 4: Calculate the Mean Value

Substitute \( t = 4 \) and \( t = 0 \) into the expression calculated in Step 3:\[\mu = \frac{3}{32} \left( \frac{4}{3}(4)^3 - \frac{1}{4}(4)^4 \right) = \frac{3}{32} \left( \frac{4}{3}(64) - 64 \right) = \frac{3}{32} \left( \frac{256}{3} - 64 \right)\]Simplify this to find the mean \( \mu \).

Step 5: Calculate the Variance

The variance \( \sigma^2 \) can be calculated by:\[\sigma^2 = \int_{0}^{4} (t^2) g(t) \, dt - \mu^2\]Evaluate the integral:\[\int_{0}^{4} t^2 \left( \frac{3}{32}(4t - t^2) \right) \, dt \]Find this value to compute the variance.

Step 6: Solve the Variance Integral

Evaluate the integral for \( t^2 \) part:\[\int_{0}^{4} t^2 \left( \frac{3}{32}(4t - t^2) \right) \, dt = \frac{3}{32} \left[ \frac{4}{4}t^4 - \frac{1}{5}t^5 \right]_{0}^{4}\]Calculate this to obtain \( \int_{0}^{4} t^2 g(t) \, dt \) and use it to find \( \sigma^2 \).

Step 7: Find Variance Value

From Step 6, calculate the integral and substitute:\[\int_{0}^{4} t^2 \left( \frac{3}{32}(4t - t^2) \right) \, dt = \frac{3}{32} \left( \frac{256}{4} - \frac{1024}{5} \right)\]Subtract the square of the mean calculated in Step 4 to find \( \sigma^2 \).

Step 8: Calculate Standard Deviation

Once variance \( \sigma^2 \) is found, take the square root to get the standard deviation \( \sigma \):\[\sigma = \sqrt{\sigma^2}\]

Step 9: Calculate Probability P(t ≤ 3)

To find \( P(t \leq 3) \), integrate the probability density function from 0 to 3:\[P(t \leq 3) = \int_{0}^{3} g(t) \, dt = \int_{0}^{3} \frac{3}{32}(4t - t^2) \, dt\]Evaluate this integral to find \( P(t \leq 3) \).

Step 10: Interpret P(t ≤ 3)

The result of \( P(t \leq 3) \) indicates the probability that a child learns the rules of the game within 3 minutes. This reflects how quickly most children in this age group understand the game.

Key Concepts

Mean Calculation

When working with a probability density function (PDF), one important measure we often need is the mean or the average value. Here, our PDF describes the learning time, in minutes, for children aged 8 to 10 learning a new board game. The mean gives us an expected learning time.
The mean of a continuous random variable is obtained by integrating the function of time, weighted by the probability for each time, over all possible times. In mathematical terms, this is represented as: \[ \mu = \int_{a}^{b} t \cdot g(t) \, dt \] Where \( g(t) \) is the probability density function, and \([a, b]\) is the range over which the function is defined, in our case 0 to 4.
Calculating the mean involves integrating the function \( g(t) = \frac{3}{32}(4t-t^{2}) \) from 0 to 4. By substituting and simplifying, we obtain the expected mean time, which tells us, on average, how long children typically take to learn the game. This is crucial for educators and game manufacturers to adjust game complexity accordingly.

Standard Deviation

After calculating the mean, another essential statistical value is the standard deviation. Standard deviation measures how spread out the learning times are around the mean.
To find the standard deviation, we first need to calculate the variance \( \sigma^2 \). The variance gives us an idea of the average squared deviation from the mean. It is calculated using: \[ \sigma^2 = \int_{a}^{b} (t - \mu)^2 \cdot g(t) \, dt \] A more common approach is to use: \[ \sigma^2 = \int_{a}^{b} t^2 \cdot g(t) \, dt - \mu^2 \] This approach leverages the calculated mean. First, integrate \( t^2 \cdot g(t) \) over 0 to 4, subtract the square of the mean from this result to get variance \( \sigma^2 \).
Finally, take the square root of \( \sigma^2 \) to find the standard deviation \( \sigma \). A smaller standard deviation indicates that most children learn at a pace close to the mean, whereas a larger one suggests a wider range of learning times. Understanding this helps tailor game rules or instructions to better cater to young learners.

Probability Interpretation

Probability plays a vital role in understanding how likely certain outcomes are. Here, we're asked to calculate the probability that a child takes 3 minutes or less to learn the game, expressed as \( P(t \leq 3) \). This is found by the cumulative distribution function (CDF), which integrates the PDF up to a specific value. To calculate this probability, integrate \( g(t) \) from 0 to 3: \[ P(t \leq 3) = \int_{0}^{3} g(t) \, dt \] By solving this integral, you find the fraction of children expected to learn the game within 3 minutes.
This interpretation is crucial: if \( P(t \leq 3) \) is high (close to 1), it means most children grasp the rules quickly, suggesting the game might be straightforward. If it's low, extra efforts may be needed to simplify instructions or provide additional learning resources to ensure young players' engagement.