Question

Breathing rates for humans can be as low as 4 breaths per minute or as high as 70 or 75 for a person doing strenuous exercise. Suppose that the resting breathing rates for college- age students have a distribution that is mound-shaped, with a mean of 14 and a standard deviation of 2.5 breaths per minute. What fraction of all students would have breathing rates in the following intervals? (a) 11.5 to 16.5 breaths per minute (b) 9 to 19 breaths per minute more than 21.5 or less than 6.5 breaths per minute

Answer 1

Solution :

Given that ,

mean = $\mu$ = 14

standard deviation = $\sigma$ = 2.5

a) P(11.5 < x < 16.5 ) = P[(11.5 - 14) / 2.5) < (x - $\mu$ ) /$\sigma$  < (16.5 - 14) / 2.5 ) ]

= P( - 1< z < 1 )

= P(z < 1 ) - P(z < - 1)

Using z table,

= 0.8413 - 0.1587

= 0.6826

b) P(9 < x < 19 ) = P[(9 - 14) / 2.5) < (x - $\mu$ ) /$\sigma$  < (19 - 14) / 2.5 ) ]

= P( - 2< z < 2)

= P(z < 2 ) - P(z < - 2)

Using z table,

= 0.9772 - 0.0228

= 0.9544

c) P(x > 21.5 or x < 6.5) = [1 - P (x - $\mu$ ) /$\sigma$ < (21.5 - 14) / 2.5) + P (x - $\mu$ ) /$\sigma$  < (6.5 - 14) / 2.5 ) ]

= 1 -P(z < 3) + P( z < -3)

Using z table

= 1 - 0.9987 + 0.0013

= 0.0013 + 0.0013

= 0.0026