Answer:
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = 0.5890 miles
Step-by-step explanation:
Step(i):-
Given mean of the Population = 100 miles per day
Given standard deviation of the Population = 23 miles per day
Let 'X' be the normal distribution
Let x₁ = 86
[tex]Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{86-100}{23} =-0.61[/tex]
Let x₂= 86
[tex]Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{125-100}{23} = 1.086[/tex]
Step(ii):-
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = P(-0.61 ≤ Z≤ 1.08)
= P(Z≤ 1.08) - P(Z≤ -0.61)
= 0.5 +A(1.08) - ( 0.5 - A(-0.61))
= A(1.08) + A(0.61) ( A(-Z)= A(Z)
= 0.3599 + 0.2291
= 0.5890
Conclusion:-
The probability that a truck drives between 86 and 125 miles in a day.
P(86≤ X≤125) = 0.5890 miles per day
