Question

The amount of fresh basil on the caprese salad at Olive Oil's follows a Normal distribution, with a mean of 1.15 ounces and a standard deviation of 0.21 ounce. A random sample of 14 plates of caprese salad is selected every day and the amount of basil is measured. What is the probability that the mean weight will exceed 1.23 ounces

Answer 1

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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(mean weight will exceed 1.23 ounces)

Sample size n = 14

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.21}{\sqrt{14}} = 0.056$

P(x > 1.23)

$P( x > 1.23) = P( z > \displaystyle\frac{1.23 - 1.15}{0.056}) = P(z > 1.428)$

$= 1 - P(z \leq 1.428)$

Calculation the value from standard normal z table, we have,  

$P(x > 1.23) = 1 - 0.923 = 0.077 = 7.7\%$

0.77 is the probability that the mean weight will exceed 1.23 ounces for a sample of 14 plates.