contestada

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 [tex]P_{60}[/tex], the 60-percentile of the distribution of temperature readings. This is the temperature reading separating the bottom 60% from the top 40%.

Respuesta :

Answer:

[tex]P_{60} = 0.254[/tex]    

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:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find [tex]P_{60}[/tex]

P(X<x) = 0.0600

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

P(X < x)  

[tex]P( X < x) = P( z < \displaystyle\frac{x - 0}{1})=0.600[/tex]  

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

[tex]\displaystyle\frac{x - 0}{1} = 0.254\\x = 0.254[/tex]

[tex]\bold{P_{60} = 0.254}[/tex]

Otras preguntas