The mean gas mileage for a hybrid car is 5656 miles per gallon. suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.23.2 miles per gallon. (a) what proportion of hybrids gets over 6161 miles per gallon? (b) what proportion of hybrids gets 5252 miles per gallon or less? left parenthesis c right parenthesis what(c) what proportion of hybrids gets between 5858 and 6161 miles per gallon? (d) what is the probability that a randomly selected hybrid gets less than 4646 miles per gallon?

To solve the questions we proceed as follows:
z=(x-μ)/σ
a] Proportion of hybrid that gets over 61 miles per gallon
z=(61-56)/3.2
z=1.5625
thus
P(X≥61)=1-P(X≤61)
P(X≤61)=0.9406
thus:
P(X≥56)=1-0.9406=0.0594

b]what proportion of hybrids gets 52 miles per gallon or​ less? 
z=(52-56)/3.2=-1.25
P(X≤52)=0.1056

c] what proportion of hybrids gets between 58 and 61 miles per​ gallon?
here we need to calculate:
P(58≤X≤61)
P(X≤61)=0.9406

P(X≤58) will be given by:
z=(58-56)/3.2=0.625
P(X≤58)=0.7357
thus:
P(58≤X≤61)=0.9406-0.7357=0.2049

d]what is the probability that a randomly selected hybrid gets less than 46 miles per​ gallon?
z=(46-56)/3.2
z=-3.125
P(X≤46)=0.0009

The correct answer would be shape of function b = 2× slope of function a.