contestada

A population of values has a normal distribution with u = 210.3 and o = 31.9. You intend to draw a randomsample of size n = 24.Find the probability that a single randomly selected value is greater than 224.6.P(X > 224.6) =Find the probablity that a sample of size n = 24 is randomly selected with a mean greater than 224.6.P(M > 224.6) =Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Respuesta :

Solution

Step 1

We first find the z score

[tex]\begin{gathered} Z=\text{ }\frac{x-\mu}{\sigma} \\ \text{where }x\text{ is the observed value} \\ \mu\text{ is mean of the sample} \\ \sigma\text{ is the standard deviation of the sample} \\ Z=\text{ }\frac{224.6-210.3}{31.9} \\ \\ Z=\frac{14.3}{31.9} \\ Z=0.448 \end{gathered}[/tex]

Step 2

P-value from Z-Table:

P(x>224.6) = 1 - P(x<224.6) = 0.3267 Four decimal places

Part B

[tex]\begin{gathered} Z=\text{ }\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ Z=\frac{224.6\text{ - 210.3}}{\frac{31.9}{\sqrt[]{24}}} \\ Z=\frac{14.3}{6.5116} \\ Z=2.196 \end{gathered}[/tex]

P(x>Z) = 0.0140 Four decimal places

Otras preguntas