Let <em>X</em> be a random variable denoting pulse rates of females as described, with mean <em>µ</em> = 74.0 bpm and standard deviation <em>σ</em> = 12.5 bpm. Then
P(70 < <em>X</em> < 78) = P((70 - <em>µ</em>)/<em>σ</em> < (<em>X</em> - <em>µ</em>)/<em>σ</em> < (78 - <em>µ</em>)/<em>σ</em>)
… = P(-0.32 < <em>Z</em> < 0.32)
… = P(<em>Z</em> < 0.32) - P(<em>Z</em> < -0.32)
… ≈ 0.6255 - 0.3745
… ≈ 0.2510
where <em>Z</em> is normally distributed with mean 0 and s.d. 1.
• • •
Next, you sample 25 females from the population and want to find the probability that their average falls between 70 and 78 bpm. Let
denote the pulse rate of the <em>i</em>-th woman from the sample, and let <em>Y</em> be the random variable for the sample mean; that is,

Recall that a sample of size <em>n</em> taken from a normal distribution with mean <em>µ</em> and s.d. <em>σ</em> has a mean that is also normally distributed with mean <em>µ</em> and s.d. <em>σ</em>/√<em>n</em>.
The mean stays the same, <em>µ</em> = 74.0, but now the s.d. is <em>σ</em>/√<em>n</em> = <em>σ</em>/5 = 2.5. So we have
P(70 < <em>Y</em> < 78) = P((70 - <em>µ</em>)/(<em>σ</em>/5) < (<em>Y</em> - <em>µ</em>)/(<em>σ</em>/5) < (78 - <em>µ</em>)/(<em>σ</em>/5))
… = P(-1.6 < <em>Z</em> < 1.6)
… = P(<em>Z</em> < 1.6) - P(<em>Z</em> < -1.6)
… ≈ 0.9452 - 0.0548
… ≈ 0.8904