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 distr…

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

`P_(60) = 0.254`

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(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(x - 0)/(1))=0.600`

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

`\displaystyle(x - 0)/(1) = 0.254\\x = 0.254`

`\bold{P_(60) = 0.254}`