Question:
The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8 cm. Find the probability that an individual distance is greater than 215.50 cm.
Answer:
The probability that an individual distance is greater than 215.50 cm is 0.1056.
Step-by-step explanation:
1. Identify the given information:
- Mean (μ) = 205.5 cm
- Standard deviation (σ) = 8 cm
- We need to find P(X > 215.50 cm)
2. Calculate the z-score for 215.50 cm:
z = (X - μ) / σ
z = (215.50 - 205.5) / 8
z = 10 / 8 = 1.25
3. The problem is now equivalent to finding P(Z > 1.25) in a standard normal distribution.
4. Using a standard normal distribution table or calculator:
P(Z < 1.25) = 0.8944
5. Since we want the probability of being greater than 1.25:
P(Z > 1.25) = 1 - P(Z < 1.25)
P(Z > 1.25) = 1 - 0.8944 = 0.1056
Additional information:
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. In this case, 215.50 cm is 1.25 standard deviations above the mean, which puts it in the upper tail of the distribution. The probability we calculated (0.1056 or about 10.56%) represents the area under the normal curve to the right of this point.