The weights of ripe watermelons grown at Mr. Smith's farm are normally distributed with a standard deviation of 2 lb. Find the mean weight of Mr. Smith's ripe watermelons if only 5% weigh less than 18 lb. (Give your answer correct to one decimal place.)
Incorrect: Your answer is incorrect. . 8.8 lb
I have tried this one several times and can not get it right, please help!!!!!
.05 is P(Z<-1.645)
That is, 18 is 1.645 std below the mean.
So, the mean is 18+2*1.645 = 21.29
To find the mean weight of Mr. Smith's ripe watermelons, we need to use the concept of Z-scores and the standard normal distribution.
Given that the standard deviation is 2 lb, we can say that 18 lb is 1.5 standard deviations below the mean. This is because the formula for the Z-score is:
Z = (X - μ) / σ
Where Z is the Z-score, X is the observed value, μ is the mean, and σ is the standard deviation.
Now, we need to find the Z-score corresponding to the 5th percentile. The 5th percentile corresponds to a Z-score of -1.645, which you can look up in a standard normal distribution table or use a calculator or statistical software.
So, we have:
-1.645 = (18 - μ) / 2
Next, we can solve the equation for the mean (μ):
-1.645 * 2 = 18 - μ
-3.29 = 18 - μ
μ = 18 - (-3.29)
μ = 21.29
Therefore, the mean weight of Mr. Smith's ripe watermelons is approximately 21.3 lb (rounded to one decimal place).
I apologize for the incorrect response you received earlier. Please let me know if you have any further questions!