\( n=18,8=4 \)
find the area of the following
a.) \( P(x<21) \)
b.) \( P(x>35) \)
c.) \( P(21
Ask by Evans Fleming. in the Philippines
Jan 22,2025
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To find the probabilities, we'll assume that the variable \( x \) follows a normal distribution with a mean \( \mu = 18 \) and a standard deviation \( \sigma = 4 \). First, we need to standardize the values using the Z-score formula: \( Z = \frac{(X - \mu)}{\sigma} \). a.) To find \( P(x < 21) \): \[ Z = \frac{(21 - 18)}{4} = 0.75 \] Using standard normal distribution tables or calculators, \( P(Z < 0.75) \) is approximately \( 0.7734 \). b.) To find \( P(x > 35) \): \[ Z = \frac{(35 - 18)}{4} = 4.25 \] Since this Z-score is very high, \( P(Z > 4.25) \) is virtually \( 0 \). c.) To find \( P(21 < x < 35) \): We already have \( P(x < 21) \) which is \( 0.7734 \) and for \( P(x < 35) \): \[ Z = \frac{(35 - 18)}{4} = 4.25 \] Thus, \( P(x < 35) \) is practically \( 1 \). Therefore, \( P(21 < x < 35) = P(x < 35) - P(x < 21) = 1 - 0.7734 = 0.2266 \). In summary: a.) \( P(x < 21) \approx 0.7734 \) b.) \( P(x > 35) \approx 0 \) c.) \( P(21 < x < 35) \approx 0.2266 \)