Assume that a set of test scores is normally distributed with a mean of 80 and a standard deviation of 15. Use the​ 68-95-99.7 rule to find the following quantities. a. The percentage of scores less than 80 is 50​%. ​(Round to one decimal place as​ needed.) b. The percentage of scores greater than 95 is 16.0​%. ​(Round to one decimal place as​ needed.) c. The percentage of scores between 50 and 95 is nothing​%.

Respuesta :

Answer:

a. Percentage of scores less than 100::::: 50%

-----------------------------------------------------

b. Relative frequency of scores less than 120:::(50+34)%

-----------------------------------------------------

c. Percentage of scores less than 140:::::(50+47.5)%

-----------------------------------------------------

d. Percentage of scores less than 80:::::::(50-34)%

--------------------------------------------------------

e. Relative frequency of scores less than 60::::(50-47.5)%

--------------------------------------------------------

f. Percentage of scores greater than 120:::(50-34)%

==========================================================

Step-by-step explanation:

The percentages will be "50.0%", "16.0%" and "81.5%". A further solution is provided below.

According to the question,

Mean,

  • 80

Standard deviation,

  • 15

(a)

The total area curve,

= 1.00

Half area under the curve,

= 0.500

then,

→ [tex]P (X< 80) = 0.500[/tex]

or,

→                   [tex]= 50.0[/tex] (%)

(b)

As we know,

→ [tex]Mean +1\times Standard \ deviation= 80+1\times 15[/tex]

                                                       [tex]= 80+15[/tex]

                                                       [tex]= 95[/tex]

According to 68-95-99.7 rule,

The remaining area of both sides will be:

= [tex]100-68[/tex]

= [tex]32[/tex] (%)

hence,

Remaining area of right side will be:

→ [tex]P(X> 95)[/tex] = [tex]\frac{32}{2}[/tex]

                    = [tex]16[/tex] (%)

(c)

We have,

  • [tex]P(X<95) = 0.16[/tex]

So,

→ [tex]P(X< 95)=1-0.16[/tex]

                    [tex]= 0.84[/tex]

Now,

→ [tex]Mean-2\times Standard \ deviation= 80-2\times 15[/tex]

                                                       [tex]=80-30[/tex]

                                                       [tex]= 50[/tex]

According to 68-95-99.7 rule,

Remaining area of both sides will be:

= [tex]1-.95[/tex]

= [tex]0.05[/tex]

Area remaining at left,

= [tex]\frac{0.05}{2}[/tex]

= [tex]0.025[/tex]

So, [tex]P(X<50) = 0.025[/tex]

hence,

→ [tex]P(50<X<95) = P(X<95)-P(X<50)[/tex]

                             [tex]= 0.84-0.025[/tex]

                             [tex]= 0.815[/tex]

or,

                             [tex]= 81.5[/tex] (%)    

Thus the above answers are correct.

Learn more:

https://brainly.com/question/15242040