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Several studies have shown a link between iron depletion without anemia (IDNA) a

ID: 2929083 • Letter: S

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 58 minutes and standard deviation 11 minutes. Training-session duration of female rowers with normal iron status is normally distributed with mean 69 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?

Explanation / Answer

Q1.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 58
standard Deviation ( sd )= 11
sample size (n) = 24
P(X > 64) = (64-58)/11/ Sqrt ( 24 )
= 6/2.245= 2.6722
= P ( Z >2.6722) From Standard Normal Table
= 0.0038

Q2.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 69
standard Deviation ( sd )= 18
sample size (n) = 24
P(X < 64) = (64-69)/18/ Sqrt ( 24 )
= -5/3.6742= -1.3608
= P ( Z <-1.3608) From Standard NOrmal Table
= 0.0868

Q3.
P(X>Y) = P( Z > (69-58) / Sqrt(18^/24 + 11^2/24) )
= P( Z > 2.55)
= 0.0054

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