The 100-meter race times at a state track meet are normally distributed with a mean of 14.62 seconds and a standard deviation of 2.13 seconds. using the standard normal probabilities table, what is the approximate probability that a runner chosen at random will have a 100-meter time less than 15.5 seconds? a.0.1894 b. 0.3409 c. 0.6591 d. 0.7910 e. 0.8106

To solve this problem, we use the t statistic. The formula for z score is:

z = (x – u) / s

where x is the sample value = 15.5 seconds or below, u is the sample mean = 14.62 seconds, s is standard dev = 2.13 seconds

z = (15.5 – 14.62) / 2.13

z = 0.41

Using standard distribution tables at z = 0.41, the value of P is:

P = 0.6591 = 65.91%

Hence there is a 65.91% chance the runner will have less than 15.5 seconds of time

c. 0.6591

f(x)= x^2 +4= 20

Step-by-step explanation:

the answer is 20 the explanation is there bye