Answer :
Answer: B . 0.9615
Step-by-step explanation:
Step-by-step explanation:
Let x be a random variable that represents the lengths of the shells .
As per given , we have
[tex]\mu=6.8[/tex] inches
[tex]\sigma=3.2[/tex] inches
n= 50
∵ [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Then for x= 6,
[tex]z=\dfrac{6-6.8}{\dfrac{3.2}{\sqrt{50}}}=-1.76776695297\approx-1.7678[/tex]
The probability the sample means would we expect to be greater than 6 inches :-
[tex]P(x>6)=P(z>-1.7678)=P(z<1.7678)\ \ [\because P(Z>-z)=P(Z<z)]\\\\=0.9614528\approx0.9615[/tex] [using the z values table]
Hence, the proportion of the sample means would we expect to be greater than 6 inches = 0.9615