Question

Let X be a continuous random variable with a normal distribution with a mean of 10 and a variance of 25. Suppose that you randomly picked 36 values for X and averaged these values together, calling this new random variable X. Find P(12X). Give your answer as a decimal rounded to four places (i.e. X.XXXX)

Answer 1

Answer:

0.0082

Step-by-step explanation:

Data provided in the question:

Mean = 10

Variance = 25

Standard deviation = $\sqrt{\text{variance}}

[tex]=\sqrt{25}=5$

n=36

Given $P(12 \leq \bar{x})$  

$=P(\bar{x} \geq 12)$  

$=P\left(\frac{\bar{x}-\mu_{x}}{\frac{\sigma}{\sqrt{n}}} \geq \frac{12-10}{\left(\frac{5}{\sqrt{36}}\right)}\right)$  

$=P(z \geqslant 2 \cdot 4)$  

$=1-P(z<2 \cdot 4)$  

By using z-table we get,

$=1-0.9918$  

= 0.0082.