Answer:
P(z>1.3) = 0.9032
Step-by-step explanation:
We are given:
Mean = 5000
Standard deviation = 1000
x = 6300
P(x>6300)=?
z-score =?
z-score = x- mean/standard deviation
z-score = 6300 - 5000/1000
z- score = 1300/1000
z-score = 1.3
So, P(x>6300) = P(z>1.3)
Looking at the z-probability distribution table and finding value:
P(z>1.3) = 0.9032
So, P(z>1.3) = 0.9032
the tempayure was at six because it was at 92c and went down to -13c then rose 7c so that means it was at 6 by 6 a.m.
Hayden Grace and Gaytar went to the store. They did not go anymore.
