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In statistics, when we talk about the expected value, we are referring to the average or mean outcome you would anticipate from a probability distribution if you could repeat the experiment a large number of times. For a binomial distribution, where you have a fixed number of trials and outcomes of either success or failure, the expected value helps in predicting how many successes you can expect in those trials.
In the given problem, we have 10,000 computers examined, with the probability of failure being 10%. The formula to find the expected value for a binomial distribution is \( E(X) = np \). This formula multiplies the number of trials \( n \) with the probability of success \( p \).

• For our example, \( n = 10,000 \) and \( p = 0.10 \), so \( E(X) = 10,000 \times 0.10 = 1,000 \).

The expected value here is 1,000. This indicates that out of the 10,000 computers, we expect 1,000 will have experienced CPU failure during the warranty period if repeated over many samples.

The standard deviation is a measure of the amount of variation or spread in a set of values. For probability distributions, it tells us how much the values tend to deviate from the expected value. When discussing a binomial distribution, the standard deviation can help understand the variability of our expected results.
For a binomial distribution, we calculate the standard deviation \( \sigma \) using the formula \( \sqrt{np(1-p)} \). This formula incorporates:

• \( n \), the number of trials,
• \( p \), the probability of success,
• and \( 1-p \) which represents the probability of failure.

Let's look at our case:

• Using \( n = 10,000 \), \( p = 0.10 \), and \( 1-p = 0.90 \), the standard deviation becomes \( \sigma = \sqrt{10,000 \times 0.10 \times 0.90} = \sqrt{900} = 30 \).

This result of 30 tells us how much the number of computers with CPU failure is likely to differ from the expected number, which is 1,000.

When dealing with large sample sizes in a binomial distribution, we often use a normal approximation. This is because calculating probabilities directly from the binomial distribution can become complex. If \( n \) is large and neither \( p \) nor \( 1-p \) is extremely close to zero, the distribution of sample proportions can be approximated by a normal distribution.
In our case, the sample size is 10,000, which is sufficiently large to use this method. We approximate the binomial distribution with the normal distribution \( N(\mu, \sigma) \), where \( \mu = np \) and \( \sigma = \sqrt{np(1-p)} \).
With \( \mu = 1,000 \) and \( \sigma = 30 \), our normal distribution becomes \( N(1,000, 30) \). This is used to find probabilities, such as the chance of having more than a certain number of failures, using tools like Z-scores for calculations.
This approximation simplifies calculations and helps predict the likelihood of different outcomes without manually crunching through complex binomial probability formulas.