The mean gas mileage for a hybrid car is 56 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a

Question

The mean gas mileage for a hybrid car is 56 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.2 miles per gallon.​

(a) What proportion of hybrids gets over 60 miles per​ gallon?

(b) What proportion of hybrids gets 52 miles per gallon or​ less?

(c )What proportion of hybrids gets between 57 and 62 miles per​ gallon?

(d) What is the probability that a randomly selected hybrid gets less than 45 miles per​ gallon?

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Rylee 2 weeks 2021-10-13T00:52:07+00:00 1 Answer 0

Answers ( )

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    2021-10-13T00:53:58+00:00

    Answer:

    a) P(X>60)=0.106

    b) P(X<52)=0.106

    c) P(57<X<62)=0.347

    d) P(X<45)=0.00029

    Step-by-step explanation:

    1) Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Let X the random variable that represent the gas mileage for a hybrid car of a population, and for this case we know the distribution for X is given by:

    X \sim N(56,3.2)  

    Where \mu=56 and \sigma=3.2

    (a) What proportion of hybrids gets over 60 miles per​ gallon?

    We are interested on this probability

    P(X>60)

    And the best way to solve this problem is using the normal standard distribution and the z score given by:

    z=\frac{x-\mu}{\sigma}

    If we apply this formula to our probability we got this:

    P(X>60)=P(z>\frac{60-56}{3.2})

    =P(z>\frac{60-56}{3.2})=P(z>1.25)

    And we can find this probability on this way:

    P(z>1.25)=1-P(z<1.25)=0.106

    (b) What proportion of hybrids gets 52 miles per gallon or​ less?

    We are interested on this probability

    P(X<52)

    And the best way to solve this problem is using the normal standard distribution and the z score given by:

    z=\frac{x-\mu}{\sigma}

    If we apply this formula to our probability we got this:

    P(X<52)=P(z<\frac{52-56}{3.2})

    =P(z<\frac{52-56}{3.2})=P(z<-1.25)

    And we can find this probability on this way:

    P(z<-1.25)=0.106

    (c )What proportion of hybrids gets between 57 and 62 miles per​ gallon?

    We are interested on this probability

    P(57<X<62)

    And the best way to solve this problem is using the normal standard distribution and the z score given by:

    z=\frac{x-\mu}{\sigma}

    If we apply this formula to our probability we got this:

    P(57<X<62)=P(\frac{57-56}{3.2}<z<\frac{62-56}{3.2})

    =P(0.3125<z<1.875)

    And we can find this probability on this way:

    P(0.3125<z<1.875)=P(z<1.875)-P(z<0.3125)=0.347

    (d) What is the probability that a randomly selected hybrid gets less than 45 miles per​ gallon?

    We are interested on this probability

    P(X<45)

    And the best way to solve this problem is using the normal standard distribution and the z score given by:

    z=\frac{x-\mu}{\sigma}

    If we apply this formula to our probability we got this:

    P(X<45)=P(z<\frac{45-56}{3.2})

    =P(z<\frac{45-56}{3.2})=P(z<-3.44)

    And we can find this probability on this way:

    P(z<-3.44)=1-P(z<-3.44)=0.00029

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