Respuesta :
Answer:
A. 491.775
B. 488.37
C. 484.55
Step-by-step explanation:
A. 100-95 = 5
thus
[tex]\frac{5}{100} = 0.05[/tex]
using the formula;
[tex]Z_{0.05} = \frac{X- \mu}{\sigma}[/tex]
where X = 500
μ = mean
σ = 5 (standard deviation)
[tex]\mu= X - Z_{0.05}\sigma[/tex]
= 500 - 1.645(5)
= 491.775
(please note that [tex]Z_{}0.05[/tex] = 1.645 from the standard normal distribution table )
B. similarly, following the above method we have
100 - 99 = 1
1/100 = 0.01
from the standard normal distribution table [tex]Z_{0.01} = 2.326[/tex]
μ = X - [tex]Z_{0.01}[/tex]σ
= 500 - 2.326(5)
= 488.37
C. 100 - 99.9 = 0.1
0.1/100 = 0.001
from the standard normal distribution [tex]Z_{0.001} = 3.09[/tex]
μ = X - [tex]Z_{0.001}[/tex]σ
= 500 - 3.09(5)
= 484.55
Part(a): The required mean is [tex]508.225[/tex]
Part(b): The required mean is [tex]511.65[/tex]
Part(b): The required mean is [tex]515.45[/tex]
Mean:
Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Let [tex]X[/tex] denotes the amount of fill-in half-liter soft drink and follows a normal distribution with parameters.
Part(a):
Calculating the mean be set to ensure a 95% of probability is,
[tex]P(X\ge500)=0.95\\1-P(Z < \frac{500-\mu}{5} )=0.95\\\frac{500-\mu}{5} =0.05\\\mu=508.225[/tex]
Part(b):
Calculating the mean be set to ensure a 99% of probability is,
[tex]P(X \ge 500)=0.99\\1-P(Z < \frac{500-\mu}{5} )=1-0.99\\\mu=511.65[/tex]
Part(c):
Calculating the mean be set to ensure a 99.9% of probability is,
[tex]P(X\ge500)=0.999\\1-P(Z < \frac{500-\mu}{5} )=0.999\\\frac{500-\mu}{5}=0.001\\ \mu=515.45[/tex]
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