The time it takes a cell to divide (called mitosis) is normally distributed with an average time of one hour and a standard deviation of 5 minutes.

Required:
a. What is the probability that a cell divides in less than 45 minutes?
b. What is the probability that it takes a cell more than 65 minutes to divide?
c. By what time have approximately 99% of all cells completed mitosis?

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