High School

In an agricultural experiment, a large, uniform field was sown with a variety of wheat. The field was divided into several plots (each plot measured 7 X 100 feet) and the harvest from each plot. the mean was 145 pounds with a standard deviation of 22 pounds. What percentage of the plots yielded 120 lbs. or more?

Answer :

The mean harvest was 145 pounds with a standard deviation of 22 pounds. We will determine the percentage of plots that yielded 120 pounds or more.

To find the percentage of plots that yielded 120 pounds or more, we need to calculate the z-score for the value of 120 pounds and then determine the area under the normal distribution curve corresponding to that z-score.

The z-score is calculated using the formula: z = (x - μ) / σ, where x is the value (120 pounds), μ is the mean (145 pounds), and σ is the standard deviation (22 pounds).

Substituting the values into the formula: z = (120 - 145) / 22 = -25 / 22 ≈ -1.14.

We can then look up the z-score of -1.14 in the standard normal distribution table or use statistical software to find the corresponding area under the curve. The area to the left of -1.14 is approximately 0.1271, which represents the percentage of plots that yielded less than 120 pounds.

To find the percentage of plots that yielded 120 pounds or more, we subtract the above percentage from 100%: 100% - 12.71% = 87.29%.

Therefore, approximately 87.29% of the plots yielded 120 pounds or more in the agricultural experiment.

Learn more about standard deviation here:

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