Given
mean of 406 grams and a standard deviation of 27 grams.
Find
The heaviest 14% of fruits weigh more than how many grams?
Explanation
given
mean = 406 gms
standard deviation = 27 gms
using standard normal table ,
![\begin{gathered} P(Z>z)=14\% \\ 1-P(Zso , [tex]\begin{gathered} x=z\times\sigma+\mu \\ x=1.08\times27+406 \\ x=435.16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%28Z%3Ez%29%3D14%5C%25%20%5C%5C%201-P%28Zso%20%2C%20%5Btex%5D%5Cbegin%7Bgathered%7D%20x%3Dz%5Ctimes%5Csigma%2B%5Cmu%20%5C%5C%20x%3D1.08%5Ctimes27%2B406%20%5C%5C%20x%3D435.16%20%5Cend%7Bgathered%7D)
Final Answer
Therefore , The heaviest 14% of fruits weigh more than 435.16 gms
I think the answer is A because the data is skewed to the right
The median (middle) isn't 10, it's 8
The interquartile range (IQR) is 7 not 6 because you have to subtract the first and third quartile 13 - 6 = 7 not 6
Hope this helps! ;)
Answer:
<h2>-1</h2>
Step-by-step explanation:
2 - (-4) + (-y)
Substitute y = 7 to the expression:
2 - (-4) + (-7) = 2 + 4 - 7 = 6 - 7 = -1
Is there a fraction or a number line to explain you question?☺
It is a right, isosceles triangle.
