A statistics practitioner determined that the mean and stan-dard deviation of a data set were 90 and 15, respectively. What can yousay about the proportions of observations that lie between each of thefollowing intervals if you know that the distribution is bell-shaped? Whatcan you say if the distribution is not bell-shaped?

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26-02-2020
Mathematics

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A statistics practitioner determined that the mean and stan-dard deviation of a data set were 90 and 15, respectively. What can yousay about the proportions of observations that lie between each of thefollowing intervals if you know that the distribution is bell-shaped? Whatcan you say if the distribution is not bell-shaped?

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dfbustos dfbustos · Answer 1 · 2020-02-27T11:22:44+01:00

Answer:

If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.

a) [tex]P(75<X<105)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(75<X<105) =P(\frac{75-90}{15} <Z<\frac{105-90}{15}) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683[/tex]

b) [tex]P(60<X<120)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(60<X<120) =P(\frac{60-90}{15} <Z<\frac{120-90}{15}) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955[/tex]

c) [tex]P(45<X<135)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(45<X<135) =P(\frac{45-90}{15} <Z<\frac{135-90}{15}) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997[/tex]

If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.

Step-by-step explanation:

Previous concepts

If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.

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".

Part a

Let X the random variable of interest

We assume for this case that [tex]\mu=90[/tex] and [tex]\sigma=15[/tex]

We are interested on this probability

[tex]P(75<X<105)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(75<X<105) =P(\frac{75-90}{15} <Z<\frac{105-90}{15}) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683[/tex]

Part b

[tex]P(60<X<120)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(60<X<120) =P(\frac{60-90}{15} <Z<\frac{120-90}{15}) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955[/tex]

Part c

[tex]P(45<X<135)[/tex]

And for this case we can use the z score given by:

[tex] z = \frac{X-\mu}{\sigma}[/tex]

And if we use it we got:

[tex] P(45<X<135) =P(\frac{45-90}{15} <Z<\frac{135-90}{15}) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997[/tex]

If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.