Final answer:
Using the z-distribution for a large sample size of 38 and given values of mean = 35.4 and standard deviation = 20.5, the 80% confidence interval for the number of seeds is approximately (31.141, 39.659). The provided options do not match this result, which suggests possible errors in the options or in the data given.
Step-by-step explanation:
To calculate the 80% confidence interval for the number of seeds in a certain fruit based on a sample mean of 35.4 and a standard deviation of 20.5 with 38 specimens, we will use the formula for a confidence interval with a t-distribution, as the sample size is less than 30:
CI = µ ± (t* × (s/√n))
Where µ is the sample mean, t* is the t-score associated with the confidence level (which can be found from a t-distribution table or calculator), s is the sample standard deviation, and n is the sample size. Since our sample size is actually greater than 30, we'll use the z-distribution. For an 80% confidence interval and a large sample size like 38, the z-score is approximately 1.282.
Now we calculate the margin of error:
Margin of Error (E) = z* × (s/√n)
= 1.282 × (20.5/√38)
≈ 1.282 × 3.325
≈ 4.259
Substitute the values into the CI formula:
CI = 35.4 ± 4.259
= (35.4 - 4.259, 35.4 + 4.259)
= (31.141, 39.659)
Thus, the 80% confidence interval for the typical number of seeds is approximately (31.141, 39.659), which is not listed in the options provided by the student. It seems there might have been an error in the options or the sample standard deviation provided could be inaccurate.