The height of 16-month-old seedlings are normally distributed with a mean of 31.5 cm and standard deviation of 10 cm. What is the height above which 75% of the seedlings have grown?
hi, ok try using the z-score tables and find 0.75 and find the z-score that corresponds. use that with the formula. except you get one value and you need a range. im not sure????
To find the height above which 75% of the seedlings have grown, we need to find the z-score associated with the 75th percentile of the standard normal distribution.
1. Start by finding the z-score using the cumulative distribution function (CDF) of the standard normal distribution. The CDF gives the probability that a value is less than or equal to a certain z-score.
- Since we want to find the height above which 75% have grown, we need to find the z-score for the 75th percentile, which corresponds to a cumulative probability of 0.75.
- We can use a standard normal distribution table or a statistical calculator to find the z-score associated with a cumulative probability of 0.75.
2. Once we have the z-score, we can use it to find the corresponding height in centimeters.
- The z-score indicates how many standard deviations the value is from the mean.
- We can use the standard formula: height = mean + (z-score * standard deviation) to calculate the height.
- In this case, the mean is 31.5 cm and the standard deviation is 10 cm.
For example, let's assume the z-score associated with the 75th percentile is 0.6745 (values are rounded for simplicity):
height = 31.5 + (0.6745 * 10)
= 38.745 cm
Therefore, the height above which 75% of the seedlings have grown is approximately 38.745 cm.