Question 1167859: 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 P81, the 81-percentile. This is the temperature reading separating the bottom 81% from the top 19%.
p81=?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! ```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.
-- end list -->
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, $P\_{81} = 0.88^\\circ \\text{C}$.
$P\_{81} = \\boxed{0.88}$
Answer by ikleyn(52779)  (Show Source):
You can put this solution on YOUR website! .
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 P81, the 81-percentile. This is the temperature reading separating the bottom 81% from the top 19%.
~~~~~~~~~~~~~~~~~~~~~~~~
To solve the problem, we should translate this set of words into meaningful human form.
We are given the normal curve with the mean value of 0 and the standard deviation of 1.
They want you find the score 'x' (the temperature) such that the area under this normal curve
on the left of 'x' be 0.81.
Use the standard function invNorm of a regular hand calculator TI-84/87.
Its format is
x = invNorm(area, mean, SD).
So, in this problem we want to calculate invNorm(0.81, 0, 0.81).
The calculator gives the ANSWER x = 0.8779.
It means that 87.79% of readings will be lower than 0.81 °C.
Alternatively, you may use the Excel function NORM.INV(area,mean,SD)
or online free of charge calculator at this web-page
https://onlinestatbook.com/2/calculators/inverse_normal_dist.html
This online calculator has an advantage showing you visually the area under the curve,
so you will learn the subject momentarily and better than if 5 or 15 teachers/tutors
will explain it to you.
With this online calculator, you will catch the problem instantly.
Solved.
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