Respuesta :
Answer:
a) 0.059
b) 0.579
c) 0.211
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 21.6 grams
Standard Deviation, σ = 6 grams
We are given that the distribution of grams of fat in a corner store purchase is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(purchase has more than 31 grams of fat)
P(x > 31)
[tex]P( x > 31) = P( z > \displaystyle\frac{31 - 21.6}{6}) = P(z > 1.567)[/tex]
[tex]= 1 - P(z \leq 1.567)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 31) = 1 - 0.941 = 0.059 = 5.9\%[/tex]
b) P(corner store purchase has between 15 and 25 grams of fat)
[tex]P(15 \leq x \leq 25) = P(\displaystyle\frac{15 - 21.6}{6} \leq z \leq \displaystyle\frac{25-21.6}{6}) = P(-1.1 \leq z \leq 0.567)\\\\= P(z \leq 0.567) - P(z < -1.1)\\= 0.715 - 0.136 = 0.579 = 57.9\%[/tex]
[tex]P(15 \leq x \leq 25) = 57.9\%[/tex]
c) Standard error due to sampling =
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{2}} = 4.24[/tex]
P(purchase has more than 25 grams of fat for 2 stores)
P(x > 25)
[tex]P( x > 25) = P( z > \displaystyle\frac{25 - 21.6}{4.24}) = P(z > 0.802)[/tex]
[tex]= 1 - P(z \leq 0.802)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 25) = 1 - 0.789 = 0.211 = 21.1\%[/tex]
