Respuesta :
Answer:
[tex]P(\bar X>3.4) = 0.385[/tex]
Step-by-step explanation:
Relevant Data provided according to the question is as follows
[tex]\mu[/tex] = 3.2
[tex]\sigma[/tex] = 0.8
n = 4
According to the given scenario the calculation of probability that the sample means will be more than 3.4 pounds is shown below:-
[tex]z = \frac{\bar X - \mu}{\frac{a}{\sqrt{n} } }[/tex]
[tex]P(\bar X>3.4) = 1 - P(\bar X\leq 3.4)[/tex]
[tex]= 1 - P \frac{\bar X - \sigma}{\frac{a}{\sqrt{n} } } \leq \frac{3.4 - \sigma}{\frac{a}\sqrt{n} }[/tex]
Now, we will solve the formula to reach the probability that is
[tex]= 1 - P \frac{\bar X - 3.2}{\frac{0.8}{\sqrt{4} } } \leq \frac{3.4 - 3.2}{\frac{0.8}\sqrt{4} }[/tex]
[tex]= 1 - P (Z \leq \frac{0.2}{0.4})[/tex]
[tex]= 1 - P (Z \leq 0.5})[/tex]
[tex]= 1 - \phi (0.5)[/tex]
= 1 - 0.6915
= 0.385
Therefore the correct answer is
[tex]P(\bar X>3.4) = 0.385[/tex]
So, for computing the probability we simply applied the above formula.
Answer:
its 21
Step-by-step explanation:
its not 21 i really dont know
