A particular fruit's weights are normally distributed, with a mean of 571 grams and a standard deviation of 18 grams.
If you pick 8 fruits at random, then 11% of the time, their mean weight will be greater than how many grams?
Give your answer to the nearest gram.
To find the weight at which the mean weight of 8 fruits is greater 11% of the time, we need to determine the z-score corresponding to the 11th percentile.
First, we need to find the z-score corresponding to the 11th percentile by using the Z-table or a calculator. The z-score corresponding to the 11th percentile is approximately -1.22.
Next, we can use the formula for the standard error of the mean to find the standard deviation of the sample distribution. The standard error of the mean is equal to the population standard deviation divided by the square root of the sample size.
Standard error of the mean = standard deviation / √(sample size)
= 18 / √(8)
= 18 / 2.828
≈ 6.36
Finally, we can find the weight at which the mean weight of 8 fruits is greater 11% of the time by multiplying the standard error of the mean by the z-score and adding it to the mean weight.
Weight = Mean weight + (z-score * standard error of the mean)
= 571 + (-1.22 * 6.36)
≈ 563.13 grams
Rounded to the nearest gram, the mean weight of 8 fruits will be greater 11% of the time when it is greater than approximately 563 grams.