Respuesta :
Answer:
a) 0.033
b) 0.468
c) 100
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 105.3
Standard Deviation, σ = 8
The amount x of miraculin produced (measured in micro-grams per gram of fresh weight) had a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(x > 120)
[tex]P( x > 120) = P( z > \displaystyle\frac{120 - 105.3}{8}) = P(z > 1.8375)[/tex]
[tex]= 1 - P(z \leq 1.8375)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 120) = 1 - 0.967 = 0.033 = 3.3\%[/tex]
b) P(x between 100 and 110)
[tex]P(100 \leq x \leq 110) = P(\displaystyle\frac{100 - 105.3}{8} \leq z \leq \displaystyle\frac{110-105.3}{8}) = P(-0.6625 \leq z \leq 0.5875)\\\\= P(z \leq 0.5875) - P(z \leq -0.6625)\\= 0.722 - 0.254 = 0.468 = 46.8\%[/tex]
[tex]P(100 \leq x \leq 110) = 46.8\%[/tex]
c) P(x < a) = 0.25
[tex]P( x \leq a) = P( z \leq \displaystyle\frac{a - 105.3}{8}) = 0.25[/tex]
Calculation the value from standard normal z table, we have,
[tex]P( z \leq -0.674) = 0.25[/tex]
[tex]\displaystyle\frac{x - 105.3}{8} = -0.674\\\\x = -0.674\times 8 + 105.3 = 99.908 \approx 100[/tex]
