Approximately what percent of women have heights between 61.5 and 66.5 inches?

Answer: 68% 61.5 and 66.5 are exactly μ ± σ. By the empirical rule about 68% of values lie within 1 SD of the mean.

Standard normal Z has

Answer: μ = 0, σ = 1 Standard normal: mean 0, SD 1.

The normal density is

Answer: Symmetric, bell-shaped Normal: symmetric about μ, bell-shaped.

AP Statistics · Topic 1.10

The Normal Distribution Practice

Part of Exploring One-Variable Data.(VAR-2.A)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$Heights of women in a population are approximately normal with μ = 64 in and σ = 2.5 in.$
$Approximately what percent of women have heights between 61.5 and 66.5 inches?$
- A
  $50%$
- B
  $99.7%$
- C
  $68%$
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- D
  $95%$
Why
$61.5 and 66.5 are exactly μ \pm σ. By the empirical rule about 68% of values lie within 1 SD of the mean.$
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Sample 2difficulty 2/5
$Standard normal Z has$
- A
  $μ = 1, σ = 1$
- B
  $μ = 0, σ = 0$
- C
  $μ = 1, σ = 0$
- D
  $μ = 0, σ = 1$
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Why
$Standard normal: mean 0, SD 1.$
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Sample 3difficulty 2/5
$The normal density is$
- A
  $Bimodal$
- B
  $Uniform$
- C
  $Symmetric, bell-shaped$
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- D
  $Skewed right$
Why
$Normal: symmetric about μ, bell-shaped.$
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Sample 4difficulty 2/5
$A density curve has total area under it equal to$
- A
  $1$
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- B
  $0$
- C
  $100$
- D
  $0.5$
Why
$Probability density integrates to 1.$
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Sample 5difficulty 2/5
$A bell-shaped, symmetric density curve is shown.$
$Which property is characteristic of normal distributions?$
- A
  $Always discrete$
- B
  $Always skewed right$
- C
  $Symmetric, bell-shaped, with mean = median$
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- D
  $Always bimodal$
Why
$Normal distributions are symmetric, bell-shaped, unimodal, and have mean equal to median.$
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