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Sagot :
Answer:
The percentage of that vehicles whose speeds are between 62 miles per hour and 70 miles per hour
P(62≤X≤70) = 68.76
Step-by-step explanation:
Step(i):-
Given that the mean of the Population = 66miles per hour
Given that the standard deviation of the Population = 4 miles per hour
Let 'X' be the random variable in normal distribution
Let x= 62
[tex]Z = \frac{x^{-}-mean }{S.D} = \frac{62-66}{4} =-1[/tex]
let x=70
[tex]Z = \frac{70-66 }{4} = \frac{70-66}{4} =1[/tex]
Step(ii):-
The probability that vehicles whose speeds are between 62 miles per hour and 70 miles per hour
P(62≤X≤70) = P(-1≤Z≤1)
= P(z≤1) -P(Z≤-1)
= P(Z≤1)+P(Z≤1)
= 2×P(Z≤1)
= 2×0.3438
= 0.6876
The percentage of that vehicles whose speeds are between 62 miles per hour and 70 miles per hour
P(62≤X≤70) = 68.76
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