Respuesta :
Answer:
a) P[ Z < 45 ] = 0,14 %
b) P[ Z > 65 ] = 16%
c) z = 71,65 minutes
Step-by-step explanation:
Normal Distribution
Mean (μ₀) 1 hour or 60 minutes
Standard Deviation σ = 5 (minutes)
a) The probability that a cell divides in less than 45 minutes is:
Let´s call Z standard normal random with distribution function Φ then:
P[ Z < 45 ] = ( Z - μ₀ ) / σ
P[ Z < 45 ] = ( 45 - 60 ) / 5
P[ Z < 45 ] = - 3
With this value we look in Z table and find the value 0,00135
P[ Z < 45 ] = 0,00135 or
P[ Z < 45 ] = 0,14 %
b)The probability of Z > 65 minutes is:
P[ Z > 65 ] = 1 - P[ Z ≤ 65 ]
P[ Z ≤ 65 ] = ( Z - μ₀ )/ σ
P[ Z ≤ 65 ] = ( 65 - 60 ) / 5
P[ Z ≤ 65 ] = 1
From Z table we find the reciprocate value of 0,84134 then
P[ Z ≤ 65 ] = 0,84134 and
P[ Z > 65 ] = 1 - 0,84134
P[ Z > 65 ] = 0,15866
P[ Z > 65 ] = 0,16 or P[ Z > 65 ] = 16%
c) The time to get 99% of all cell completed the process is:
P [ Z ≤ z ] = ( z - 60 ) / σ
In Z table the value of 0,99010 which is a good aproximation of 0,99 (or 99%) reciprocate with the value 2,33 then
P [ Z ≤ z ] = 2,33 = ( z - 60 ) / σ
2,33 * 5 = z - 60
11,65 + 60 = z
z = 71,65 minutes
As a matter of fact checking this value, we can see according to the empirical rule that 99,7 % of all values will occur
mean + 3σ or 60 + 15 = 75
In this exercise we have to use statistical knowledge to calculate the probability of cells dividing, so we can say that:
a) [tex]P[ Z < 45 ] = 0,14 \%[/tex]
b) [tex]P[ Z > 65 ] = 16\%[/tex]
c) [tex]Z = 71,65 minutes[/tex]
With the knowledge and information given about probability we have:
a) The probability that a cell divides in less than 45 minutes, so let´s call Z standard normal random with distribution function then:
[tex]P[ Z < 45 ] = ( Z - \mu_0 ) / \sigma\\P[ Z < 45 ] = ( 45 - 60 ) / 5\\P[ Z < 45 ] = - 3\\P[ Z < 45 ] = 0,00135 \ or \ P[ Z < 45 ] = 0,14 \%[/tex]
b)The probability of Z > 65 minutes is:
[tex]P[ Z > 65 ] = 1 - P[ Z \leq 65 ]\\P[ Z \leq 65 ] = ( Z - \mu_0 )/ \sigma\\P[ Z \leq 65 ] = ( 65 - 60 ) / 5\\P[ Z \leq 65 ] = 1\\P[ Z \leq 65 ] = 0,84134 \ and \ P[ Z > 65 ] = 1 - 0,84134\\P[ Z > 65 ] = 0,15866\\P[ Z > 65 ] = 0,16 \ or \ P[ Z > 65 ] = 16\%[/tex]
c) The time to get 99% of all cell completed the process is:
[tex]P [ Z\leq z ] = ( z - 60 ) / \sigma\\P [ Z \leq z ] = 2,33 = ( z - 60 ) / \sigma\\2,33 * 5 = z - 60\\11,65 + 60 = z\\z = 71,65 minutes[/tex]
See more about stattistics at brainly.com/question/10951564