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Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is [tex]$P(x \geq 92)$[/tex]?
To determine the probability \(P(x \geq 92)\) for a normal distribution with a mean of 98 and a standard deviation of 6, follow these steps:
1. Identify the given parameters: - Mean (\(\mu\)) = 98 - Standard deviation (\(\sigma\)) = 6 - Value of \(x\) = 92
2. Compute the z-score: The z-score helps us determine how many standard deviations away \(x\) is from the mean. The formula for the z-score is: [tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex] Plugging in the given values: [tex]\[
z = \frac{92 - 98}{6} = \frac{-6}{6} = -1
\][/tex]
3. Find the cumulative probability corresponding to the z-score: Using standard normal distribution tables or a cumulative distribution function (CDF) calculator, we find the cumulative probability for \(z = -1\). This cumulative probability \(P(X < 92)\) is approximately 0.1587.
4. Calculate \(P(x \geq 92)\): The probability \(P(x \geq 92)\) is the complement of \(P(X < 92)\): [tex]\[
P(x \geq 92) = 1 - P(X < 92)
\][/tex] Substituting the cumulative probability: [tex]\[
P(x \geq 92) = 1 - 0.1587 = 0.8413
\][/tex]
Therefore, the probability \(P(x \geq 92)\) is approximately 0.8413.
Among the options provided, the closest value is: C. 0.84
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