Questions: The lifetime of a 2-volt non-rechargeable battery in constant use has a Normal distribution, with a mean of 516 hours and a standard deviation of 20 hours. The proportion of batteries with lifetimes exceeding 520 hours is approximately

Solution Steps

To find the Z-score for a battery lifetime of 520 hours, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 520 \)
• \( \mu = 516 \)
• \( \sigma = 20 \)

Substituting the values, we have:

\[ z = \frac{520 - 516}{20} = \frac{4}{20} = 0.2 \]

Thus, the Z-score for 520 hours is \( z = 0.2 \).

Next, we need to find the probability that a battery lasts longer than 520 hours, which can be expressed as:

\[ P(X > 520) = P(Z > 0.2) \]

Using the cumulative distribution function \( \Phi \), we can express this as:

\[ P(X > 520) = \Phi(\infty) - \Phi(0.2) \]

Since \( \Phi(\infty) = 1 \), we have:

\[ P(X > 520) = 1 - \Phi(0.2) \]

From standard normal distribution tables or calculations, we find:

\[ \Phi(0.2) \approx 0.5793 \]

Thus, the probability becomes:

\[ P(X > 520) = 1 - 0.5793 = 0.4207 \]

Final Answer