A population is normally distributed with a mean of 43.2 and a standard deviation of 5.7. Find the following probabilities. (Round your answers to four decimal places.) (a) P(43.2 < x < 48.9) (b) P(41.4 < x < 44.6) (c) P(x < 50.0) (d) P(33.8 < x < 52.6) (e) P(x = 45.8) (f) P(x > 45.8)
The Correct Answer and Explanation is:
To solve these problems, we need to use the Z-score formula and the standard normal distribution table. The Z-score formula is:Z=X−μσZ = \frac{X – \mu}{\sigma}Z=σX−μ
where:
- XXX is the value of interest,
- μ\muμ is the mean (43.2),
- σ\sigmaσ is the standard deviation (5.7).
We will then use the Z-scores to find the probabilities using the cumulative distribution function (CDF) of the standard normal distribution.
(a) P(43.2 < x < 48.9)
First, calculate the Z-scores for 43.2 and 48.9:
For X=43.2X = 43.2X=43.2:Z1=43.2−43.25.7=0Z_1 = \frac{43.2 – 43.2}{5.7} = 0Z1=5.743.2−43.2=0
For X=48.9X = 48.9X=48.9:Z2=48.9−43.25.7=1Z_2 = \frac{48.9 – 43.2}{5.7} = 1Z2=5.748.9−43.2=1
Now, find the corresponding cumulative probabilities:
- P(Z=0)=0.5000P(Z = 0) = 0.5000P(Z=0)=0.5000 (from standard normal table)
- P(Z=1)=0.8413P(Z = 1) = 0.8413P(Z=1)=0.8413
Thus:P(43.2<x<48.9)=P(Z2)−P(Z1)=0.8413−0.5000=0.3413P(43.2 < x < 48.9) = P(Z_2) – P(Z_1) = 0.8413 – 0.5000 = 0.3413P(43.2<x<48.9)=P(Z2)−P(Z1)=0.8413−0.5000=0.3413
(b) P(41.4 < x < 44.6)
For X=41.4X = 41.4X=41.4:Z1=41.4−43.25.7=−0.3158Z_1 = \frac{41.4 – 43.2}{5.7} = -0.3158Z1=5.741.4−43.2=−0.3158
For X=44.6X = 44.6X=44.6:Z2=44.6−43.25.7=0.2456Z_2 = \frac{44.6 – 43.2}{5.7} = 0.2456Z2=5.744.6−43.2=0.2456
From the standard normal table:
- P(Z=−0.3158)≈0.3762P(Z = -0.3158) \approx 0.3762P(Z=−0.3158)≈0.3762
- P(Z=0.2456)≈0.5968P(Z = 0.2456) \approx 0.5968P(Z=0.2456)≈0.5968
Thus:P(41.4<x<44.6)=P(Z2)−P(Z1)=0.5968−0.3762=0.2206P(41.4 < x < 44.6) = P(Z_2) – P(Z_1) = 0.5968 – 0.3762 = 0.2206P(41.4<x<44.6)=P(Z2)−P(Z1)=0.5968−0.3762=0.2206
(c) P(x < 50.0)
For X=50.0X = 50.0X=50.0:Z=50.0−43.25.7=1.1263Z = \frac{50.0 – 43.2}{5.7} = 1.1263Z=5.750.0−43.2=1.1263
From the standard normal table:P(Z=1.1263)≈0.8708P(Z = 1.1263) \approx 0.8708P(Z=1.1263)≈0.8708
Thus:P(x<50.0)=0.8708P(x < 50.0) = 0.8708P(x<50.0)=0.8708
(d) P(33.8 < x < 52.6)
For X=33.8X = 33.8X=33.8:Z1=33.8−43.25.7=−1.6526Z_1 = \frac{33.8 – 43.2}{5.7} = -1.6526Z1=5.733.8−43.2=−1.6526
For X=52.6X = 52.6X=52.6:Z2=52.6−43.25.7=1.6579Z_2 = \frac{52.6 – 43.2}{5.7} = 1.6579Z2=5.752.6−43.2=1.6579
From the standard normal table:
- P(Z=−1.6526)≈0.0495P(Z = -1.6526) \approx 0.0495P(Z=−1.6526)≈0.0495
- P(Z=1.6579)≈0.9521P(Z = 1.6579) \approx 0.9521P(Z=1.6579)≈0.9521
Thus:P(33.8<x<52.6)=P(Z2)−P(Z1)=0.9521−0.0495=0.9026P(33.8 < x < 52.6) = P(Z_2) – P(Z_1) = 0.9521 – 0.0495 = 0.9026P(33.8<x<52.6)=P(Z2)−P(Z1)=0.9521−0.0495=0.9026
(e) P(x = 45.8)
Since the probability of a single point in a continuous distribution is always 0:P(x=45.8)=0P(x = 45.8) = 0P(x=45.8)=0
(f) P(x > 45.8)
For X=45.8X = 45.8X=45.8:Z=45.8−43.25.7=0.4561Z = \frac{45.8 – 43.2}{5.7} = 0.4561Z=5.745.8−43.2=0.4561
From the standard normal table:P(Z=0.4561)≈0.6756P(Z = 0.4561) \approx 0.6756P(Z=0.4561)≈0.6756
Thus:P(x>45.8)=1−P(Z=0.4561)=1−0.6756=0.3244P(x > 45.8) = 1 – P(Z = 0.4561) = 1 – 0.6756 = 0.3244P(x>45.8)=1−P(Z=0.4561)=1−0.6756=0.3244
Summary of Results:
- (a) P(43.2 < x < 48.9) = 0.3413
- (b) P(41.4 < x < 44.6) = 0.2206
- (c) P(x < 50.0) = 0.8708
- (d) P(33.8 < x < 52.6) = 0.9026
- (e) P(x = 45.8) = 0
- (f) P(x > 45.8) = 0.3244
