Question

Several studies have shown a link between iron depletion without anemia (IDNA) and physical performance. In one recent study, the physical performance of 24 female collegiate rowers with IDNA was compared with 24 female collegiate rowers with normal iron status. Several different measures of physical performance were studied, but we'll focus here on training-session duration. Assume that training-session duration of female rowers with IDNA is normally distributed with mean 59 minutes and standard deviation 11 minutes. Training-session duration of female rowers with normal iron status is normally distributed with mean 68 minutes and standard deviation 18 minutes.

(a) What is the probability that the mean duration of the 24 rowers with IDNA exceeds 64 minutes?


(b) What is the probability that the mean duration of the 24 rowers with normal iron status is less than 64 minutes?


(c) What is the probability that the mean duration of the 24 rowers with IDNA is greater than the mean duration of the 24 rowers with normal iron status?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	59
std deviation   =σ=	11.0000
sample size       =n=	24
std error=σx̅=σ/√n=	2.2454

probability that the mean duration of the 24 rowers with IDNA exceeds 64 minutes :

probability =

P(X>64)

=

P(Z>2.23)=

1-P(Z<2.23)=

1-0.9871=

0.0129

b)

here mean=       μ=	68
std deviation   =σ=	18.0000
sample size       =n=	24
std error=σx̅=σ/√n=	3.6742

probability that the mean duration of the 24 rowers with normal iron status is less than 64 minutes :

probability =

P(X<64)

=

P(Z<-1.09)=

0.1379

c)

mean difference =59-68=-9

and std deviation of difference =sqrt(11²/24+18²/24) =4.306

hence probability that the mean duration of the 24 rowers with IDNA is greater than the mean duration of the 24 rowers with normal iron status =P(Xbar>0)=P(Z>(0-(-9))/4.306)=P(Z>2.09)=0.0183