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Sagot :
To determine the probability [tex]\( P(x \geq 92) \)[/tex] for a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 98 and a standard deviation ([tex]\(\sigma\)[/tex]) of 6, follow these steps:
1. Calculate the z-score for [tex]\( x = 92 \)[/tex]:
The z-score formula is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
Substituting the given values:
[tex]\[ z = \frac{92 - 98}{6} \][/tex]
[tex]\[ z = \frac{-6}{6} \][/tex]
[tex]\[ z = -1 \][/tex]
2. Find the cumulative probability [tex]\( P(Z < -1) \)[/tex]:
The cumulative distribution function (CDF) of a standard normal distribution provides the probability that a standard normal variable [tex]\( Z \)[/tex] is less than a specified value.
For [tex]\( z = -1 \)[/tex]:
[tex]\[ P(Z < -1) \approx 0.1587 \][/tex]
3. Determine [tex]\( P(x \geq 92) \)[/tex]:
To find [tex]\( P(x \geq 92) \)[/tex], we need to calculate the probability that [tex]\( X \)[/tex] is greater than or equal to 92. This is the complement of [tex]\( P(X < 92) \)[/tex]:
[tex]\[ P(X \geq 92) = 1 - P(X < 92) \][/tex]
From the previous step, we know:
[tex]\[ P(X < 92) \approx 0.1587 \][/tex]
Therefore:
[tex]\[ P(X \geq 92) = 1 - 0.1587 = 0.8413 \][/tex]
So the probability [tex]\( P(x \geq 92) \)[/tex] is approximately 0.8413, which corresponds to option C (rounded to two decimal places).
Answer: C. 0.84
1. Calculate the z-score for [tex]\( x = 92 \)[/tex]:
The z-score formula is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
Substituting the given values:
[tex]\[ z = \frac{92 - 98}{6} \][/tex]
[tex]\[ z = \frac{-6}{6} \][/tex]
[tex]\[ z = -1 \][/tex]
2. Find the cumulative probability [tex]\( P(Z < -1) \)[/tex]:
The cumulative distribution function (CDF) of a standard normal distribution provides the probability that a standard normal variable [tex]\( Z \)[/tex] is less than a specified value.
For [tex]\( z = -1 \)[/tex]:
[tex]\[ P(Z < -1) \approx 0.1587 \][/tex]
3. Determine [tex]\( P(x \geq 92) \)[/tex]:
To find [tex]\( P(x \geq 92) \)[/tex], we need to calculate the probability that [tex]\( X \)[/tex] is greater than or equal to 92. This is the complement of [tex]\( P(X < 92) \)[/tex]:
[tex]\[ P(X \geq 92) = 1 - P(X < 92) \][/tex]
From the previous step, we know:
[tex]\[ P(X < 92) \approx 0.1587 \][/tex]
Therefore:
[tex]\[ P(X \geq 92) = 1 - 0.1587 = 0.8413 \][/tex]
So the probability [tex]\( P(x \geq 92) \)[/tex] is approximately 0.8413, which corresponds to option C (rounded to two decimal places).
Answer: C. 0.84
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