## Statistics Undergrad Course

Question

Determine the area under the standard normal curve that lies…….

A) To the left of -2.34 or to the right of 645.

B) Between -0.88 and 55

C) To the left of -1 or to the right of 2

D) Between 1.48 and 2.82

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Data Interpretation
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## Answers ( 3 )

Hi there! Thanks for asking this question.

To start with the standard normal curve, the values are found out with a table called

Biometrika.However, various universities/books/authors/courses use different tables. Just wanted to know which table you are using.I am using this table:

Basic Concept: Middle value is 0 and areas on each side are 0.5, so total area 1 under the complete curve.

## Example calculations:

From the table, the area from 0 to 1 is 0.3413.

And, the area to the left of 0 is 0.5.

So, total area to the left of 1 is 0.5 + 0.3413 = 0.8413

Negative values need assistance from their positive counterparts.

-1 has its counterpart 1.

To the left of -1, lies the same area which lies to the right of 1.

From the table, the area between 0 and 1 = 0.3413

So, the area beyond 1 = 0.5 – 0.3413 = 0.1587

So, this will be the required area to the left 0f -1.

Now let’s come to the first two parts of your question.

645 and 55are values which aretoo highfor standard normal distribution. Theyalmost touch the line and the area beyond those doesn’t matter.This is so because 99.27% area under the normal curve lies between -3 and 3.## Your answers are here:

A.To the left of -2.34:Means, to the right of 2.34.

From the table, the area between 0 and 2.34 = 0.4904

So, the area to the right of 2.34 = 0.5 – 0.4904 = 0.0096

So, required area is 0.0096

To the right of 645: Almost zero area.B.Between -0.88 and 55:Means to the right of -0.88, because the area to the right of 55 would be negligible, so counting it or not doesn’t matter.

The area to the right of -0.88 = the area to the left of 0.88

= 0.5 + the area between 0 and 0.88

= 0.5 + 0.3106

= 0.8106

C.To the left of -1: (described in the example above) = 0.1587To the right of 2:From the table, the area between 0 and 2 = 0.4772So, the area to the right of 2 = 0.5 – 0.4772 = 0.0228

D.Between 1.48 and 2.82:(i) The area between 0 and 1.48 (From table) = 0.4306

(ii) The area between 0 and 2.82 (From table) = 0.4976

So, the area between 1.48 and 2.82 = 0.4976 – 0.4306 = 0.0670

I MADE A MISTAKE ON (A) AND (B) 🙁

(A) SHOULD SAY 1.645 INSTEAD OF 645

(B) SHOULD SAY 2.55 INSTEAD OF 55

MY APOLOGIES 🙁