Answer: 0.2487
Step-by-step explanation:
Given: Mean: [tex]\mu=16.38[/tex] meters
Standard deviation: [tex]\sigma= 1.34[/tex] meters
Let X denote the distance in the triple jump.
The probability that a randomly selected distance from the distribution would fall into interval 16.8 meters and 17.9 meters = [tex]P(16.8<X<17.9)=P(\dfrac{16.8-16.38}{1.34}<\dfrac{X-\mu}{\sigma}<\dfrac{17.9-16.38}{1.34})[/tex]
[tex]=P(0.3134<Z<1.1343)\ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=P(Z<1.1343)-P(Z<0.3134)\\\\=0.8717- 0.6230\ [\text{By z-table}]\\\\=0.2487[/tex]
Hence, the probability that a randomly selected distance from the distribution would fall into interval 16.8 meters and 17.9 meters= 0.2487
