Respuesta :
Answer:
395.55 grams should be specified to ensure that 95 percent of the boxes will contain more than this weight.
Explanation:
We are given the following information in the question:
Mean, μ = 412 grams
Standard Deviation, σ = 10 grams
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.95
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 412}{10})=0.95[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 412}{10})=0.95 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 412}{10})=0.05 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 412}{10} = -1.645\\\\x = 395.55[/tex]
Hence, 395.55 grams should be specified to ensure that 95 percent of the boxes will contain more than this weight.
