Question 1167859
```python?code_reference&code_event_index=2
from scipy.stats import norm

# Given parameters
mean = 0  # degrees Celsius
std_dev = 1.00  # degrees Celsius
percentile = 81  # as a percentage

# Convert percentile to a decimal for the PPF function
probability = percentile / 100

# Find the Z-score corresponding to the 81st percentile
# norm.ppf(q, loc=0, scale=1) returns the inverse of the CDF, i.e., the value x
# such that P(X <= x) = q, for a normal distribution with mean=loc and std_dev=scale.
z_score_p81 = norm.ppf(probability)

# Convert the Z-score back to the temperature reading (x)
# x = mean + Z * std_dev
p81_temperature = mean + z_score_p81 * std_dev

print(f"The Z-score for the 81st percentile is: {z_score_p81:.3f}")
print(f"P81, the 81-percentile temperature reading is: {p81_temperature:.2f}°C")
```

```text?code_stdout&code_event_index=2
The Z-score for the 81st percentile is: 0.878
P81, the 81-percentile temperature reading is: 0.88°C

```

To find $P\_{81}$, the 81st percentile temperature reading, we use the properties of the normal distribution.

Given:

  * Mean ($\\mu$) = $0^\\circ \\text{C}$
  * Standard Deviation ($\\sigma$) = $1.00^\\circ \\text{C}$
  * Percentile = 81st percentile, which means we are looking for the temperature reading ($x$) such that $81%$ of the readings are below it.

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1.  **Find the Z-score corresponding to the 81st percentile:**
    The Z-score is the value on the standard normal distribution (mean = 0, standard deviation = 1) that corresponds to a cumulative probability of 0.81.
    Using a Z-table or statistical software, the Z-score for $P(Z \\le z) = 0.81$ is approximately $0.878$.

2.  **Convert the Z-score back to the temperature reading:**
    We use the formula: $x = \\mu + Z\\sigma$
    $x = 0 + (0.878) \\times 1.00$
    $x = 0.878$

Rounding to two decimal places, $P\_{81} = 0.88^\\circ \\text{C}$.

$P\_{81} = \\boxed{0.88}$