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.)
lb
Can someone help with the formula and I want to attempt to work it out.
I have worked several times this is my last try and I got (18-.5199)/2=8.74005=8.7
Z = (score-mean)/SD = (18-mean)/2
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) and its Z score. Insert in equation above and solve for mean.
When I worked out this is what I came up with..(31.55-31.5)/0.003= 16.66666 = 16.6667
Sure! To find the mean weight of Mr. Smith's ripe watermelons, we need to use the concept of the z-score and the standard normal distribution.
The formula we will use is:
z = (x - μ) / σ
Where:
z is the z-score,
x is the observed value (18 lb),
μ is the mean weight of the watermelons, and
σ is the standard deviation (2 lb).
In this case, we want to find the mean weight, μ, such that only 5% of the watermelons weigh less than 18 lb.
To determine the z-score corresponding to a 5% lower tail area, we use a standard normal distribution table or calculator.
The z-score that corresponds to a 5% lower tail area is approximately -1.645.
Now we can solve for μ using the formula:
-1.645 = (18 - μ) / 2
To isolate μ, we multiply both sides by 2 and rearrange the equation:
-3.29 = 18 - μ
Next, we isolate μ by subtracting 18 from both sides:
-3.29 - 18 = -μ
-21.29 = -μ
Finally, we multiply both sides by -1 to solve for μ:
21.29 = μ
So, the mean weight of Mr. Smith's ripe watermelons is approximately 21.3 lb.