In 16 one-hour test runs, the gasoline consumption of an engine averaged 16.4 gallons with a standard deviation of 2.1. Based on historical data, the expected gasoline consumption of this engine is...

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In 16 one-hour test runs, the gasoline consumption of an engine averaged 16.4 gallons with a standard deviation of 2.1. Based on historical data, the expected gasoline consumption of this engine is 12.0 gallons per hour (μ" role="presentation" style="display: inline-block; line-height: normal; text-align: left; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">μ). Find the probability that our sample of 16 one-hour test runs is within a gallon of the expected gasoline consumption.That is, what isP(11≤X¯≤13)" role="presentation" style="display: inline-block; line-height: normal; text-align: left; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">P(11≤X¯≤13)P(11

Answered Same DayApr 26, 2022

Answer To: In 16 one-hour test runs, the gasoline consumption of an engine averaged 16.4 gallons with a...

Pooja answered on Apr 26 2022
108 Votes
mean=    16.40
sd= sqrt(var)    2.1000
        
I know that, z = (X-mean)/(sd)    
z1 = (11-16.4)/2.1) =    -2.571
4
z2 = (13-16.4)/2.1) =    -1.6190
    
hence,    
P(11 < X < 13)    
= P(X<13) - P(X<11)    
= P(Z<-1.619) - P(Z<-2.5714)    
= NORMSDIST(-1.619) - NORMSDIST(-2.5714)    
= 0.0477
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