How do I solve this?: The times of all 15 year olds who run a certain race are approximately normally distributed with a given m
ean = 18 sec and standard deviation = 1.2 sec. What percentage of the runners have times less than 14.4 sec?
2 answers:
We need to 'standardise' the value of X = 14.4 by first calculating the z-score then look up on the z-table for the p-value (which is the probability)
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%
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The full question is attached in the picture below.
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Step-by-step explanation:
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Then, we can conclude than the 6 pack gives us a better price per battery
