contestada

Different dealers may sell the same car for different prices. The sale prices for a particular car are normally distributed with a mean and standard deviation of 26 thousand dollars and 2 thousand dollars, respectively. Suppose we select one of these cars at random. Let X represent the sale price (in thousands of dollars) for the selected car.1. Find P(26< X<30) You may round your answer to two decimal places

Respuesta :

Answer:

0.48

Step-by-step explanation:

P(26<X<30)=?

This probability can be calculated using normal distribution.

Here mean=μ=26

standard deviation=σ=2

So, P(26<X<30)=P(26-26/2<X-μ/σ<30-26/2)

P(26<X<30)=P(0<Z<2)

P(26<X<30)=0.4772=0.48

Hence P(26<X<30)=0.48=48%

MrRoyal

Probabilities are used to determine the chances of events

The probability that P(26<x<30) is 0.48

The given parameters are:

  • [tex]\mu = 26[/tex] -- the mean
  • [tex]\sigma = 2[/tex] -- the standard deviation

The probability P(26<X<30) is represented as:

[tex]P(26<x<30)=P(z1 <z<z2)[/tex]

Where:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]P(26<x<30)=P(\frac{26-26}{2}<z<\frac{30-26}{2})[/tex]

Simplify

[tex]P(26<x<30)=P(\frac{0}{2}<z<\frac{4}{2})[/tex]

Further, simplify

[tex]P(26<x<30)=P(0<z<2)[/tex]

Rewrite the above equation as:

[tex]P(26<x<30)=P(z <2) - P(z<0)[/tex]

For z-score of probabilities, we have:

[tex]P(26<x<30)=0.97725-0.5[/tex]

Simplify

[tex]P(26<x<30)=0.47725[/tex]

Approximate

[tex]P(26<x<30)=0.48[/tex]

Hence, the value of the probability that P(26<x<30) is 0.48

Read more about probabilities at:

https://brainly.com/question/15246027

Otras preguntas