brainly A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.5 years, and standard deviation of 1.2

Question

brainly A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.5 years, and standard deviation of 1.2 years. If 24 items are picked at random, 8% of the time their mean life will be less than how many years?

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Eva 2 weeks 2021-10-01T20:56:56+00:00 1 Answer 0

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    2021-10-01T20:58:14+00:00

    Answer:

    z=-1.405<\frac{a-5.5}{1.2}

    And if we solve for a we got

    a=5.5 -1.405*1.2=3.814

    So the value of interest that separates the bottom 8% of data from the top 72% is 69.764.  

    Step-by-step explanation:

    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”.  

    Solution to the problem

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

    X \sim N(5.5,1.2)  

    Where \mu=5.5 and \sigma=1.2

    For this part we want to find a value a, such that we satisfy this condition:

    P(X>a)=0.92   (a)

    P(X<a)=0.08   (b)

    Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

    As we can see on the figure attached the z value that satisfy the condition with 0.08 of the area on the left and 0.92 of the area on the right it’s z=-1.405. On this case P(Z<-1.405)=0.08 and P(z>-1.405)=0.5

    If we use condition (b) from previous we have this:

    P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.08  

    P(z<\frac{a-\mu}{\sigma})=0.08

    But we know which value of z satisfy the previous equation so then we can do this:

    z=-1.405<\frac{a-5.5}{1.2}

    And if we solve for a we got

    a=5.5 -1.405*1.2=3.814

    So the value of height that separates the bottom 8% of data from the top 72% is 69.764.  

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