Respuesta :
Answer:
0.77 is the probability that the mean weight will exceed 1.23 ounces for a sample of 14 plates.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 1.15 ounces
Standard Deviation, σ = 0.21 ounce
We are given that the distribution of amount of fresh basil is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(mean weight will exceed 1.23 ounces)
Sample size n = 14
Standard error due to sampling =
[tex]\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.21}{\sqrt{14}} = 0.056[/tex]
P(x > 1.23)
[tex]P( x > 1.23) = P( z > \displaystyle\frac{1.23 - 1.15}{0.056}) = P(z > 1.428)[/tex]
[tex]= 1 - P(z \leq 1.428)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 1.23) = 1 - 0.923 = 0.077 = 7.7\%[/tex]
0.77 is the probability that the mean weight will exceed 1.23 ounces for a sample of 14 plates.
