The distribution of time needed to complete a certain programming task is approximately normal, with mean 47 minutes and standard deviation 6 minutes. What is the probability that a randomly chosen task will take less than 34 minutes or more than 60 minutes to complete?

To identify the probability that a randomly chosen task will take less than 34 minutes or more than 60 minutes to complete, we need to calculate z-score and p-values of the 34 min. and 60 min.

That is P(X<34) + P(X>60)=P(z<z(34)) + P(z>z(60))

z score can be calculated using the formula

z=[tex]\frac{X-M}{s}[/tex] where

X is the value we are looking for its z-score (34,60)

M is the mean time to complete a certain programming task (47 min.)

s is the standard deviation (6 min.)

Then z-score of 34 min is:

z(34)=[tex]\frac{34-47}{6}[/tex] ≈ -2.17

and P(z<z(34)) =0.015

z-score of 60 min is:

z(34)=[tex]\frac{60-47}{6}[/tex] ≈ 2.17

and P(z<z(60)) =0.985, P(z>z(60)) = 1-0.985=0.015

Thus

the probability that a randomly chosen task will take less than 34 minutes or more than 60 minutes to complete is 0.015+0.015=0.03

shubheetutor shubheetutor · Answer 2 · 2025-03-25T11:22:39-04:00

The probability that a randomly chosen task will take less than [tex]34[/tex] minutes or more than [tex]60[/tex] minutes to complete is [tex]0.03[/tex].

To identify the probability that a randomly chosen task will take less than [tex]34[/tex] minutes or more than [tex]60[/tex] minutes to complete, we need to calculate the z-score and P (Probability) of the [tex]34\;\rm{minutes}[/tex] and [tex]60\;\rm{minutes}[/tex] i.e., [tex]P(X<34) + P(X>60)=P(z<z(34)) + P(z>z(60))[/tex]

z-score is calculated by using the formula, [tex]Z=\dfrac{X-\mu}{\sigma}[/tex] where,

X is the raw score i.e., [tex](34 , 60)[/tex]

[tex]\mu[/tex] is the meantime to complete a certain programming task ([tex]47 min[/tex])

[tex]\sigma[/tex] is the standard deviation ([tex]6 min[/tex])

Then, z-score of [tex]34 min[/tex] is calculated as:

[tex]Z=\dfrac{34-47}{6}\\Z=-2.17[/tex]

and, [tex]P(z<z(34)) =0.015[/tex]

z-score of [tex]60 min[/tex] is:

[tex]Z=\dfrac{60-47}{6}\\Z=2.17[/tex]

and

[tex]P(z<z(60)) =0.985\\P(z>z(60)) = 1-0.985\\P(z>z(60)) =0.015[/tex]

Hence,

[tex]\rm{Total\;Probability}=0.015+0.015\\\rm{Total\;Probability}=0.03[/tex]

The probability that a randomly chosen task will take less than [tex]34 \;\rm minutes[/tex] or more than [tex]60\;\rm minutes[/tex] to complete is 0.03.

Learn more about probability distribution here:

https://brainly.com/question/14210034?referrer=searchResults