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The times of all 15 year olds who run a certain race are approximately normally distributed with a given mean mc026-1.jpg = 18 sec and standard deviation mc026-2.jpg = 1.2 sec. What percentage of the runners have times less than 14.4 sec?
0.15%
0.30%
0.60%
2.50%

Respuesta :

InesWalston

Answer:

A.  0.15%

Step-by-step explanation:

The times of all 15 year old who run a certain race are approximately normally distributed with a given mean as 18 sec and standard deviation as 1.2 sec.

We know that,

[tex]z=\dfrac{X-\mu}{\sigma}[/tex]

where,

X = raw score = 14.4

μ = mean = 18

σ = standard deviation = 1.2

Putting the values,

[tex]z=\dfrac{14.4-18}{1.2}=\dfrac{-3.6}{1.2}=-3[/tex]

Finding the the probability from z score table, we get

[tex]P(z<-3)=0.00135=0.135\%[/tex]

ajinems

The percentage of runners that have times less than 14.4 seconds that is P(x<14.4) is 0.0013499 which is approximately = 0.15%.

What is standard deviation?

The standard deviation of a data set is defined as how the data is dispersed in relation to the mean.

From the question,

The raw time given (X) = 14.4 sec

The mean value(m) = 18sec

The standard deviation(d) = 1.2 sec

Using the formula,

z = X - m/ d

z= 14.4- 18/1.2

z = -3.6/1.2

z= -3

Using a Z table to find the percentage equivalent of P(x<14.4) is 0.0013499 which is approximately = 0.15%.

Therefore, the percentage of runners that have times less than 14.4 seconds that is P(x<14.4) is 0.0013499 which is approximately = 0.15%.

Learn more about standard deviation here:

https://brainly.com/question/475676

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