Questions: Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P72, the 72-percentile. This is the temperature reading separating the bottom 72% from the top 28%.
P72= °C
(Round answer to three decimal places)
Transcript text: Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of $0^{\circ} \mathrm{C}$ and a standard deviation of $1.00^{\circ} \mathrm{C}$. A single thermometer is randomly selected and tested. Find $P_{72}$, the 72 -percentile. This is the temperature reading separating the bottom $72 \%$ from the top $28 \%$.
$P_{72}=$ $\square$ ${ }^{\circ} \mathrm{C}$
(Round answer to three decimal places)
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Solution
Solution Steps
Step 1: Determine the Z-Score for the 72nd Percentile
To find the 72nd percentile (\(P_{72}\)) of a normally distributed dataset, we first calculate the z-score corresponding to this percentile. The z-score for the 72nd percentile is given by:
\[
z_{0.72} \approx 0.5828
\]
Step 2: Convert the Z-Score to the Original Scale
Next, we convert the z-score back to the original temperature scale using the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the distribution. The formula to convert the z-score to the original value is:
\[
P_{72} = \mu + z_{0.72} \cdot \sigma
\]
Substituting the values:
\[
P_{72} = 0 + 0.5828 \cdot 1 = 0.5828
\]
Step 3: Round the Result
Finally, we round the result to three decimal places: