Answer:
93.0% of the cans will have 10.82 oz of soda or more.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12 oz
Standard Deviation, σ = 0.8 oz
We are given that the distribution of soda fills is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.93.
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 12}{0.8})=0.93[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 12}{0.8})=0.93 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 12}{0.8})=0.07 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z<-1.476) = 0.07[/tex]
[tex]\displaystyle\frac{x - 12}{0.8} = -1.476\\x = 10.8192 \approx 10.82[/tex]
Hence, 93.0% of the cans will have 10.82 oz of soda or more.
