Many snails have a 1 mile race; the time it takes for them to complete the race

Question

Many snails have a 1 mile race; the time it takes for them to complete the race is normally distributed with a mean of 50 hours and standard deviation of 6 hours.

What is the percentage of snails that take more than 60 hours to finish?
What is the relative frequency of snails that would take less than 60 hours to finish?
What is the proportion of snails that would take between 60 and 67 hours to finish?
What is the probability that a random snail would take more than 76 hours to finish?
What is the most hours a snail must finish to be among the 10% fastest snails?
The typical 80% snails take between what two hours to finish?

Explanation / Answer

population mean,u = 50
standard deviation,sigma = 6

a.
proportion of snails that take more than 60 hours to finish is
P(X> 60 ) = P(Z> ((60-50) / 6)
= P(Z> 1.67 )
= 1 - P(Z<1.67)
= 1 - 0.9525
= 0.0475
= 4.75 %

To find P(Z<1.67) look in the standard normal table in the row 1.60 and column 0.07 (1.67 = 1.60 + 0.07), the value obtained is 0.9525

b.
relative frequency of snails that take less than 60 hours to finish is
P(X< 60 ) = P(Z< ((60-50) / 6)
= P(Z< 1.67 )
= 0.9525

c.
proportion of snails that take between 60 and 67 hours to finish is
P( 60 <X< 67 )
= P( ((60-50) / 6) <Z< ((67-50) / 6) )
= P( 1.67 <Z< 2.83 )
= P(Z<2.83) - P(Z<1.67)
= 0.9977 - 0.9525
= 0.0452

d.
P(X> 76 ) = P(Z> ((76-50) / 6)
= P(Z> 4.33 )
= 1 - P(Z<4.33)
= 1 - 0.9999
= 0.0001

e.
P(X>k) = 0.10
P(X<k) = 0.90
P(Z<(k-50)/6) = 0.90

from table 0.90 = P(Z<1.28)
(k-50)/6 = 1.28
k = 57.68

f.
P(-z<Z<z) = 0.80
2*P(Z<z) - 1 = 0.80
P(Z<z) = 0.90
z = 1.28

lower bound = u - 1.28*sigma = 42.32
upper bound = u + 1.28*sigma = 57.68

most typical 80% of snails take between 42.32 and 57.68 hours to finish

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