Answer:
The answer is below
Step-by-step explanation:
The weights of ice cream cartons produced by a manufacturer are normally distributed with a mean weight of 10 ounces and a standard deviation of 0.5 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 10.21 ounces? (b) You randomly select 25 cartons. What is the probability that their mean weight is greater than 10.21 ounces
Answer:
Given that:
Mean (μ) = 10 ounces, standard deviation (σ) = 0.5 ounces.
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score (z) is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\For\ a\ sample\ size(n):\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]
a) For x = 10.21:
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{10.21-10}{0.5}=0.42[/tex]
From the normal distribution table, probability that a randomly selected carton has a weight greater than 10.21 ounces = P(x > 10.21) = P(z > 0.42) = 1 - P(z < 0.42) = 1 - 0.6628 = 0.3372
b ) For x = 10.21 and n = 25
[tex]\sqrt{x} \sqrt{x} z=\frac{x-\mu}{\sigma/\sqrt{n} }\\\\z=\frac{10.21-10}{0.5/\sqrt{25 } }=2.1[/tex]
From the normal distribution table, probability that a randomly selected carton has a weight greater than 10.21 ounces = P(x > 10.21) = P(z > 2.1) = 1 - P(z < 2.1) = 1 - 0.9826 = 0.0174
