Answer: [tex]\mu=15.64[/tex]
Step-by-step explanation:
Let x be a random variable that represent the amount dispensed by a soft-drink dispensing machine which has a normal distribution with mean μ and standard deviation [tex]\sigma=[/tex] 0.155 ounce.
To find : μ such that P(X>16)=0.01
Consider P(X>16)=0.01
[tex]P(\dfrac{X-\mu}{\sigma}>\dfrac{16-\mu}{0.155})=0.01[/tex]
Since, [tex]Z=\dfrac{X-\mu}{\sigma}[/tex]
Also, z-table , one -tailed z-value for p-value of 0.01= 2.326
Then, [tex]\dfrac{16-\mu}{0.155}=2.326[/tex]
[tex]\Rightarrow {16-\mu}=0.155\times2.326\\\\\Rightarrow {16-\mu}=0.36053\\\\\Rightarrow\ \mu=16-0.36053=15.63947\approx15.64\\\\\Rightarrow\ \mu=15.64[/tex]