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"Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for metal sheets of a particular type, its mean value and standard deviation are 75 GPa and 2.1 GPa, respectively. Suppose the distribution is normal. (Round your answers to four decimal places.)(a) Calculate P(74 ? X ? 76) when n = 25.(b)How likely is it that the sample mean diameter exceeds 76 when n = 49?"

Respuesta :

Answer:

a)[tex]P(74 <\bar X <76)=P(-2.38< Z< 2.38) = P(Z<2.38)-P(Z<-2.38)=0.9913-0.0087=0.9827[/tex]

b)[tex]P(\bar X >76)=P(Z>3.33)=1-P(Z<3.33)=1-0.9996=0.0004[/tex]

So on this case is very unlikely that the sample mean would be higher exceeds 76 when n =49

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

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

[tex]X \sim N(75,2.1)[/tex]  

Where [tex]\mu=75[/tex] and [tex]\sigma=2.1[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu=75, \frac{\sigma}{\sqrt{n}}=\frac{2.1}{\sqrt{25}})[/tex]

And we want to find this probability:

[tex]P(74 <\bar X <76)=P(\frac{74-75}{\frac{2.1}{\sqrt{25}}}<Z<\frac{76-75}{\frac{2.1}{\sqrt{25}}})[/tex]

[tex]P(-2.38< Z< 2.38) = P(Z<2.38)-P(Z<-2.38)=0.9913-0.0087=0.9827[/tex]

3) Part b

For this case the new distribution is given by:

[tex]\bar X \sim N(\mu=75, \frac{\sigma}{\sqrt{n}}=\frac{2.1}{\sqrt{49}})[/tex]

[tex]P(\bar X >76)=P(Z>\frac{76-75}{\frac{2.1}{\sqrt{49}}}=3.33)[/tex]

[tex]P(Z>3.33)=1-P(Z<3.33)=1-0.9996=0.0004[/tex]

So on this case is very unlikely that the sample mean would be higher exceeds 76 when n =49

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