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The 5th percentile of a standard normal distribution is -1.645 (from the normal probability table).

z = (x - μ) / σ

-1.645 = (x-98.2) / .62

(0.62)(-1.645) = x-98.2

-1.0199 = x-98.2

x = 98.2-1.0199 = 97.1801

The 5th percentile is 97.18

The 95th percentile of a standard normal distribution is 1.645 (from the normal probability table).

z = (x - μ) / σ

1.645 = (x-98.2) / .62

(0.62)(1.645) = x-98.2

1.0199 = x-98.2

x = 98.2+1.0199 = 99.2199

The 95th percentile is 99.22

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Q: What is the fifth and 9fifth percentiles of a mean temperature of 98.2 and the standard deviation is 0.62?
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