Questions: Use the given values of n=1205 and p=0.98 to find the maximum value that is significantly low, μ-2σ, and the minimum value that is significantly high, μ+2σ. Round your answer to the nearest hundredth unless otherwise noted.

Solution Steps

To solve this problem, we need to calculate the mean ($\mu$) and the standard deviation ($\sigma$) for a binomial distribution. The formulas for these are:

• $\mu = n \cdot p$
• $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Once we have $\mu$ and $\sigma$, we can find the maximum value that is significantly low ($\mu - 2\sigma$) and the minimum value that is significantly high ($\mu + 2\sigma$).

1. Calculate the mean ($\mu$) using the formula $\mu = n \cdot p$.
2. Calculate the standard deviation ($\sigma$) using the formula $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$.
3. Compute $\mu - 2\sigma$ for the maximum value that is significantly low.
4. Compute $\mu + 2\sigma$ for the minimum value that is significantly high.
5. Round the results to the nearest hundredth.

The mean (\(\mu\)) for a binomial distribution is given by: \[ \mu = n \cdot p \] Substituting the given values \(n = 1205\) and \(p = 0.98\): \[ \mu = 1205 \cdot 0.98 = 1180.9 \]

The standard deviation (\(\sigma\)) for a binomial distribution is given by: \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \] Substituting the given values \(n = 1205\) and \(p = 0.98\): \[ \sigma = \sqrt{1205 \cdot 0.98 \cdot (1 - 0.98)} = \sqrt{1205 \cdot 0.98 \cdot 0.02} \approx 4.8598 \]

The maximum value that is significantly low is given by: \[ \mu - 2\sigma \] Substituting the calculated values \(\mu = 1180.9\) and \(\sigma \approx 4.8598\): \[ \mu - 2\sigma = 1180.9 - 2 \cdot 4.8598 \approx 1171.18 \]

The minimum value that is significantly high is given by: \[ \mu + 2\sigma \] Substituting the calculated values \(\mu = 1180.9\) and \(\sigma \approx 4.8598\): \[ \mu + 2\sigma = 1180.9 + 2 \cdot 4.8598 \approx 1190.62 \]

Final Answer