To find the probability that the selected monthly payment is greater than \$1400, we first calculate the Z-score for \( X = 1400 \):
Z=X−μσ=1400−982180≈2.3222 Z = \frac{X - \mu}{\sigma} = \frac{1400 - 982}{180} \approx 2.3222 Z=σX−μ=1801400−982≈2.3222
Using the standard normal distribution, we find:
P(X>1400)=1−P(Z<2.3222)≈1−0.9899=0.0101 P(X > 1400) = 1 - P(Z < 2.3222) \approx 1 - 0.9899 = 0.0101 P(X>1400)=1−P(Z<2.3222)≈1−0.9899=0.0101
Next, we calculate the probability that the selected monthly payment is between \$650 and \$975. We find the Z-scores for both values:
For X=650 X = 650 X=650:
Z650=650−982180≈−1.8444 Z_{650} = \frac{650 - 982}{180} \approx -1.8444 Z650=180650−982≈−1.8444
For X=975 X = 975 X=975:
Z975=975−982180≈−0.0389 Z_{975} = \frac{975 - 982}{180} \approx -0.0389 Z975=180975−982≈−0.0389
Now, we can find the probability:
P(650<X<975)=P(Z<−0.0389)−P(Z<−1.8444)≈0.4841−0.0322=0.4519 P(650 < X < 975) = P(Z < -0.0389) - P(Z < -1.8444) \approx 0.4841 - 0.0322 = 0.4519 P(650<X<975)=P(Z<−0.0389)−P(Z<−1.8444)≈0.4841−0.0322=0.4519
Thus, the final answers are:
P(X>1400)≈0.0101 \boxed{P(X > 1400) \approx 0.0101} P(X>1400)≈0.0101 P(650<X<975)≈0.4519 \boxed{P(650 < X < 975) \approx 0.4519} P(650<X<975)≈0.4519
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