In certain experiment the error made in determining the density of a substance is a random
variable having Uniform density with α = -0.015 and β=0.015. Find the probability that
(i) such an error will be between -0.01 and 0.02. and (ii) such an error will exceed 0.005.

Given that the error in determining the density of a substance is uniformly distributed with \( \alpha = -0.015 \) and \( \beta = 0.015 \), we can use the properties of the **Uniform Distribution** to solve the problem.

### Key points:
- For a uniform distribution \( U(\alpha, \beta) \), the probability density function (PDF) is given by:

\[
f(x) = \frac{1}{\beta - \alpha}, \quad \text{for} \quad \alpha \leq x \leq \beta
\]

In this case:
\[
\alpha = -0.015 \quad \text{and} \quad \beta = 0.015
\]
So the PDF is:
\[
f(x) = \frac{1}{0.015 - (-0.015)} = \frac{1}{0.03} = \frac{1}{30}
\]
Thus, the uniform probability density is constant over the interval \( [-0.015, 0.015] \).

### (i) The probability that the error is between \(-0.01\) and \(0.02\):

We need to find the probability that the error lies between \(-0.01\) and \(0.02\). Since \(0.02\) exceeds \( \beta = 0.015 \), we only consider the range from \(-0.01\) to \(0.015\).

The probability is the area under the PDF from \( x = -0.01 \) to \( x = 0.015 \):

\[
P(-0.01 \leq X \leq 0.015) = \frac{\text{Length of the interval}}{\beta - \alpha} = \frac{0.015 - (-0.01)}{0.03} = \frac{0.025}{0.03} = \frac{5}{6}
\]

Thus, the probability is:

\[
P(-0.01 \leq X \leq 0.015) = \frac{5}{6} \approx 0.8333
\]

### (ii) The probability that the error exceeds \(0.005\):

We need to find the probability that the error is greater than \(0.005\). This corresponds to the interval \( [0.005, 0.015] \).

The probability is the area under the PDF from \( x = 0.005 \) to \( x = 0.015 \):

\[
P(X > 0.005) = \frac{0.015 - 0.005}{0.03} = \frac{0.01}{0.03} = \frac{1}{3}
\]

Thus, the probability is:

\[
P(X > 0.005) = \frac{1}{3} \approx 0.3333
\]

### Summary of results:
1. \( P(-0.01 \leq X \leq 0.015) = \frac{5}{6} \approx 0.8333 \)
2. \( P(X > 0.005) = \frac{1}{3} \approx 0.3333 \)

Would you like further details on uniform distribution or any other aspect of this problem?

#### Related questions:
1. How do you compute probabilities for a uniform distribution in general?
2. What is the expected value of a uniformly distributed random variable?
3. How do you calculate variance for a uniform distribution?
4. Can you explain how uniform distribution applies to real-world problems?
5. How does the cumulative distribution function (CDF) for a uniform distribution work?

**Tip**: When working with uniform distributions, probabilities can often be computed easily using simple proportions.

In certain experiment the error made in determining the density of a substance is a random variable having Uniform density with α = -0.015 and β=0.015. Find the probability that (i) such an error will be between -0.01 and 0.02. and (ii) such an error will exceed 0.005.

Explore how to calculate the probability of errors in a uniform distribution with boundaries α = -0.015 and β = 0.015. This problem involves finding probabilities for ranges such as -0.01 to 0.02 and values exceeding 0.005. Learn the key formulas and concepts, ideal for Grades 11-12 students.

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