Question

Assume that the octane level of mid-grade gasoline has a normal distribution with a mean of 89.15 and a standard deviation of 0.16. We take many samples of this gasoline and measure their octane levels.

Problem 1: What proportion of the samples will have octane levels higher than 89.50?
Solution: Let XX denote the octane level, which is normal with μ=89.15μ=89.15 and σ=0.16σ=0.16. We need to find the probability that XX is greater than 89.5089.50.

P(X>89.50)=P(X−μσ>89.50−89.150.16)=P(Z>2.19)P(X>89.50)=P(X−μσ>89.50−89.150.16)=P(Z>2.19)

Many normal probability tables give "cumulative probabilities." Therefore, we must take the complement.

P(Z>2.19)=1−P(Z≤2.19)=1−0.9857=P(Z>2.19)=1−P(Z≤2.19)=1−0.9857=  (Round to 4 decimal places.)

Problem 2: Above what value of octane level will you find 69.5% of the samples?

Solution: First, we need to find a zz-score such that 30.5% of the area under the normal curve is found below that value. In other words, we are looking at a zz-score that gives cumulative probability 0.3050. According to the table, this number is z=−0.51z=−0.51.

Next, we need to convert this zz-score into xx. Recall z=x−μσz=x−μσ. Solving this equation for xx gives x=μ+zσx=μ+zσ. Therefore, the final answer is

x=89.15+(−0.51)(0.16)=x=89.15+(−0.51)(0.16)=  (Round to 2 decimal places.)

Answer 1

1. Let X denote the octane level, Then X follows Normal (89.15, 0.16). We need to find P(X > 89.5).Then,
, Where Z follows Normal(0 , 1)

2. First, we need to find a z-score such that 30.5% of the area under the normal curve is found below that value. In other words, we are looking at a z-score that gives cumulative probability 0.3050. According to the table,

Next, we need to convert this z-score into x. Recall that

.

Solving this equation for x we get x= -0.51*0.16 + 89.15 = 89.07