The Heights Of Women Aged 20 To 29 Are Approximately Normal

The Heights Of Women Aged 20 To 29 Are Approximately Normal Wit

The heights of women aged 20 to 29 are approximately normal with a mean of 64 inches and a standard deviation of 2.7 inches. Similarly, men of the same age group have a mean height of 69.3 inches and a standard deviation of 2.8 inches. The following calculations involve determining z-scores for specific heights, using standard normal distribution tables to find proportions, and assessing probabilities related to manufacturing specification and SAT scores.

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Calculating z-scores for individual heights involves subtracting the mean from the observed value and dividing by the standard deviation. For example, a woman who is 5'8" tall (which is 68 inches) has a z-score calculated as (68 - 64) / 2.7 ≈ 1.48. Similarly, a man who is 5'1" tall (61 inches) has a z-score of (61 - 69.3) / 2.8 ≈ -2.89. These z-scores indicate how many standard deviations these heights are from the respective population means.

Using Table A (the standard normal distribution table), we can find the proportions of observations below or above specific z-scores:

  • For z
  • For z > 0.94, the proportion is 1 - 0.8264 = 0.1736 (17.36%).
  • For z > 0.6, the proportion is approximately 1 - 0.7257 = 0.2743 (27.43%).
  • For 0.6

Finding the z-values associated with specific cumulative proportions involves referring to inverse standard normal calculations:

  • For the 10% lower tail, z ≈ -1.28 (since approximately 10% of observations fall below z = -1.28).
  • For the median (50%), z = 0, because the normal distribution is symmetric.

To find the z-value such that 10.04% of the distribution falls below z, we consult the z-table and approximate z ≈ -1.25. Similarly, to find z such that 57.62% of observations are greater than z, equivalently 42.38% are less than z, resulting in z ≈ -0.19, as the cumulative probability at z = -0.19 is approximately 42.38%.

In manufacturing, the width of slots cut by a milling machine follow an N(0.905, 0.001) distribution, with tolerance limits set between 0.90475 and 0.90525 inches. To find the proportion of slots meeting specifications, we compute the z-scores for these limits:

  • Lower limit: (0.90475 - 0.905)/0.001 = -0.25
  • Upper limit: (0.90525 - 0.905)/0.001 = 0.25

Using the standard normal distribution table, the cumulative probability at z = ±0.25 is approximately 0.5987 and 0.4013, respectively. The proportion of slots within the specification limits is approximately 0.5987 - 0.4013 = 0.1974, or 19.74%.

Next, the height comparison states that young women have heights modeled as N(64, 2.58), and young men follow N(69.3, 2.68). To find what percent of young men are shorter than the mean height of young women, we calculate the z-score for 64 inches in the men's distribution: 

z = (64 - 69.3) / 2.8 ≈ -1.89

Using the z-table, the cumulative probability at z = -1.89 is approximately 0.0294, or 2.94%. Thus, about 2.94% of young men are shorter than the mean height of young women.

Addressing SAT scores, in 2007 the distribution for men's scores is N(533, 116). The proportion scoring 760 or above involves calculating the z-score: 

z = (760 - 533) / 116 ≈ 1.94

Using the z-table, the cumulative probability up to z = 1.94 is approximately 0.9744, meaning the proportion of men scoring 760 or more is 1 - 0.9744 = 0.0256, or 2.56%.

For women with scores following N(499, 110), the z-score for 760 is: 

z = (760 - 499) / 110 ≈ 2.41

The cumulative probability at z = 2.41 is approximately 0.9920, so about 1 - 0.9920 = 0.008 or 0.8% of women scored 760 or higher in 2007.

References

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