In: Math
he 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
5151
miles per gallon or less?
left parenthesis c right parenthesis What(c) What
proportion of hybrids gets between
5959
and
6161
miles per gallon? (d) What is the probability that a randomly selected hybrid gets less than
4646
miles per gallon?
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(a) The proportion of hybrids that gets over
6161
miles per gallon is
nothing.
(Round to four decimal places as needed.)
(b) The proportion of hybrids that gets
5151
miles per gallon or less is
nothing.
(Round to four decimal places as needed.)
(c) The proportion of hybrids that gets between
5959
and
6161
miles per gallon is
nothing.
(Round to four decimal places as needed.)
(d) The probability that a randomly selected hybrid gets less than
4646
miles per gallon is
nothing
a)
µ = 56
σ = 3.2
P ( X ≥ 61 ) = P( (X-µ)/σ ≥ (61-56) /
3.2)
= P(Z ≥ 1.563 ) = P( Z <
-1.563 ) = 0.0591
(answer)
b)
µ = 56
σ = 3.2
P( X ≤ 51 ) = P( (X-µ)/σ ≤ (51-56)
/3.2)
=P(Z ≤ -1.563 ) =
0.0591 (answer)
c)
µ = 56
σ = 3.2
we need to calculate probability for ,
P ( 59 < X <
61 )
=P( (59-56)/3.2 < (X-µ)/σ < (61-56)/3.2 )
P ( 0.938 < Z <
1.563 )
= P ( Z < 1.563 ) - P ( Z
< 0.938 ) =
0.9409 - 0.8257 =
0.1152
d)
µ = 56
σ = 3.2
P( X < 46 ) = P( (X-µ)/σ < (46-56)
/3.2)
=P(Z < -3.125 ) = 0.0009
(answer)