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the heights of adult men in america are normally distributed, with a me…

Question

the heights of adult men in america are normally distributed, with a mean of 69.5 inches and a standard deviation of 2.63 inches. the heights of adult women in america are also normally distributed, but with a mean of 64.4 inches and a standard deviation of 2.53 inches.
a) if a man is 6 feet 3 inches tall, what is his z - score (to two decimal places)?
z=

b) if a woman is 5 feet 11 inches tall, what is her z - score (to two decimal places)?
z=

c) who is relatively taller?
○ the 5 foot 11 inch american woman
○ the 6 foot 3 inch american man
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Explanation:

Step1: Convert height to inches

  • For part a: 6 feet 3 inches = \(6\times12 + 3=75\) inches.
  • For part b: 5 feet 11 inches = \(5\times12+11 = 71\) inches.

Step2: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\)

  • Part a:
  • Given \(\mu = 69.5\) (mean for men), \(\sigma=2.63\) (standard deviation for men), and \(x = 75\).
  • \(z=\frac{75 - 69.5}{2.63}=\frac{5.5}{2.63}\approx2.09\)
  • Part b:
  • Given \(\mu = 64.4\) (mean for women), \(\sigma = 2.53\) (standard deviation for women), and \(x = 71\).
  • \(z=\frac{71-64.4}{2.53}=\frac{6.6}{2.53}\approx2.61\)

Step3: Compare z - scores

  • The z - score of the woman (\(z\approx2.61\)) is greater than the z - score of the man (\(z\approx2.09\))

Answer:

a) \(z\approx2.09\)
b) \(z\approx2.61\)
c) The 5 - foot 11 - inch American woman