Answer :
To solve this problem, we'll use the properties of the normal distribution. The mean yield for a one-acre plot is given as [tex]\mu = 562[/tex] kilos and the standard deviation as [tex]\sigma = 32[/tex] kilos.
(i) Yield Over 700 Kilos:
First, determine the z-score for 700 kilos using the formula:
[tex]z = \frac{X - \mu}{\sigma}[/tex]
Substitute the given values:
[tex]z = \frac{700 - 562}{32} = \frac{138}{32} \approx 4.31[/tex]
Using the z-table, a z-score of 4.31 is extremely high, much higher than typical values found within z-tables (often only up to 3.49), resulting in a probability very close to 0. This implies that practically no plots would exceed 700 kilos in yield; statistically, it would be less than one plot.
(ii) Yield Below 650 Kilos:
Calculate the z-score for 650 kilos:
[tex]z = \frac{650 - 562}{32} = \frac{88}{32} = 2.75[/tex]
Using a z-table, a z-score of 2.75 corresponds to a probability of about 0.9960. This implies that around 99.60% of the plots would have a yield below 650 kilos. Hence, in a batch of 1,000 plots, you would expect:
[tex]0.9960 \times 1000 \approx 996 \text{ plots}[/tex]
(iii) Lowest Yield of the Best 100 Plots:
To find the lowest yield of the best 100 plots, determine the top 10% yield (as 100 plots is 10% of 1,000 plots). The top 10% corresponds to the 90th percentile.
The z-score for the 90th percentile is approximately 1.28. Use the z-score formula to find the corresponding yield:
[tex]X = \mu + z \times \sigma[/tex]
[tex]X = 562 + 1.28 \times 32[/tex]
[tex]X = 562 + 40.96 \approx 603[/tex] kilos
Thus, the lowest yield among the top 100 plots is approximately 603 kilos.
