Answer: d) 0.31
Step-by-step explanation:
Given : In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes.
i.e. [tex]\mu=42.5[/tex] and [tex]\sigma= 5.4[/tex]
It is assumed that this is a normally distributed variable.
Let x denotes the time spend by a person on treadmill.
Then, the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.
[tex]P(30<x<40)=P(\dfrac{30-42.5}{5.4}<\dfrac{x-\mu}{\sigma}<\dfrac{40-42.5}{5.4})\\\\\approxP(-2.31<z<0.46)\\\\=P(z<-0.46)-P(z<-2.31)\ \ [\because\ P(z_1<z<z_2)=P(z<z_2)-P(z<z_1)]\\\\=1-P(z<0.46)-(1-P(z\leq2.31))\ \ [\because P(Z>z)=1-P(Z\leq z)]\\\\=1-0.6772419-(1-0.9895559)\ \ \text{[By using z-table or calculator]}\\\\=0.3227581-0.0104441=0.312314\approx0.31[/tex]
Hence, the required probability = 0.31
Thus , the correct answer = d) 0.31
