Question

The average length of wheat seeds follows the normal curve with an average of 6.15 mm and an SD of 0.27 mm.

What percentage of seeds are more than 6 mm in length?

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Answer 1

To find the percentage of seeds that are more than 6 mm in length, you can calculate the z-score and use a standard normal distribution table or calculator.

Step-by-step explanation:

To find the percentage of seeds that are more than 6 mm in length, we need to calculate the z-score for 6 mm using the given mean and standard deviation. The z-score formula is:

z = (x - mean) / standard deviation

Substituting the values, we get:

z = (6 - 6.15) / 0.27 = -0.5556

We can then use a standard normal distribution table or a calculator to find the percentage of seeds with a z-score greater than -0.5556.

Based on the z-score, we can determine that approximately 29.17% of the seeds are longer than 6 mm in length

Answer 2

Step-by-step explanation:

We need the z-score for 6 mm in order to find the percentage. The z-score is:

`z=(6.15-6)/(.27)` which is the mean minus the number in question all divided by the standard deviation. This division gives us .55 and the value for this z-score is .7088 which in a percent is 71%