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Fed [463]
2 years ago
13

A large random sample was taken of body temperatures of women at a university. The data from the sample were normally distribute

d with a mean body temperature of
98.52°F and a standard deviation of 0.727°F. Based on this sample, which percentage is the best estimate of the proportion of all women at this university who have a body
temperature more than 2 standard deviations above the mean?
O 0.30%
O 2.28%
72.70%
O 97.72%
Mathematics
1 answer:
ivanzaharov [21]2 years ago
6 0

The best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Since the sample is normally distributed and has a mean μ = 98.52 F and a standard deviation, σ = 0.727 F and we need to find the percentage of all womenat this university who have a body temperature more than 2 standards deviations above the mean.

<h3>The normal distribution</h3>

Since the sample is normally distributed, 50% of the sample is below the mean. We have 34% of the sample at 1 standard deviation away from the mean and 47¹/₂% at 2 standard deviations above the mean.

<h3>Percentage below 2 standard deviations away from mean</h3>

So, the percentage below 2 standard deviations from the mean is 50% + 47¹/₂% = 97¹/₂%.

<h3>Percentage above 2 standard deviations away from mean</h3>

So, the percentage above 2 standard deviations from the mean is 100% - 97¹/₂% = 2¹/₂% = 2.5 %

Since 2.28 % is the closest to 2.5 % from the options, the best estimate is 2.28 %.

So, the best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Learn more about normal distribution here:

brainly.com/question/25800303

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Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

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1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

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∠GDE = 60° (Given)

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\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

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∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

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60° = 2 × ∠CFE

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∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

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