The height of baby tree are normally distributed with a mean of 31.5 cm and a standard deviation of 10 cm. What is the height above which 85% of the seedlings have grown?
To find the height above which 85% of the seedlings have grown, we need to find the corresponding z-score using the standard normal distribution.
The first step is to convert the given percentage (85%) into a decimal probability (0.85), as the standard normal distribution uses decimal probabilities.
Now, we need to find the z-score that corresponds to a cumulative probability of 0.85. We can use a standard normal distribution table or a calculator to find this value. The z-score corresponding to a cumulative probability of 0.85 is approximately 1.036.
Next, we can use the formula for z-score transformation to find out the actual value of the height. The formula is:
z = (x - μ) / σ
Where:
z is the z-score
x is the observed value (the height we want to find)
μ is the mean
σ is the standard deviation
Rearranging the formula, we get:
x = (z * σ) + μ
Plugging in the values, we get:
x = (1.036 * 10) + 31.5
Calculating this, we find:
x = 41.36 + 31.5
x ≈ 72.86 cm
Therefore, the height above which 85% of the seedlings have grown is approximately 72.86 cm.