Question

Assume that the readings at freezing on a batch of thermometers are Normally distributed with mean 0°C and standard deviation 1.00°C. Find $P_{60}$, the 60-percentile of the distribution of temperature readings. This is the temperature reading separating the bottom 60% from the top 40%.

Answer 1

Answer:

$P_{60} = 0.254$    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0

Standard Deviation, σ = 1.00

We are given that the distribution of readings at freezing on a batch of thermometers is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find $P_{60}$

P(X<x) = 0.0600

We have to find the value of x such that the probability is 0.600

P(X < x)  

$P( X < x) = P( z < \displaystyle\frac{x - 0}{1})=0.600$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 0}{1} = 0.254\\x = 0.254$

$\bold{P_{60} = 0.254}$