The 200-meter race times at a state track meet are normally distributed with a mean of 13.56 seconds and a standard deviation of 2.24 seconds. using the standard normal probabilities table, what is the approximate probability that a runner chosen at random will have a 200-meter time less than 13.5 seconds?

P(X< 13.5) = 0.834

Step-by-step explanation:

Given data:

mean of race time is

standard deviation is

we know that z is given as

where x is 13.5 in given problem

so probability of having runner time less than 13.5

P(X< 13.5) = P(z< -0.026)

P(X< 13.5) = P(z>(1 -0.026))

P(X< 13.5) = P(z > 0.974)

from standard z table , for z = 0.974 we have 0.834

P(X< 13.5) = 0.834

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

y=mx+