suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed 4.5 hours. what would you expect the average time allowed the 15 selected students to be? (round your answer to two decimal places.)



Answer :

The average time allowed the 15 selected students to be 3.06hours

What is Average?

In plain English, an average is a single number chosen to represent a group of numbers; it is often the sum of the numbers divided by the number of numbers in the group. The average of the numbers 2, 3, 4, 7, and 9 is, for instance, 5.

With a sample size of 15 and a p value of 0.04

From binomial probability, (nCx)px(1p) is where the application of binomial probability originates from (n-x)

1 - p = 0.96

a) P(x=1) is the probability that precisely 1 received a special accommodation.

= 15C1 x (0.04)^1 x (0.96) ^14 = 0.3388

b) P(x >=1) = 1 - P(x = 0) is the likelihood that at least 1 person received a special accommodation.

= 1 - [ 15C0 x (0.04)^0 x (0.96)^15 = 0.4579

c) the likelihood that two people at least obtained special accommodations

= [P(x = 0) + P(x = 1)] - [P(x>=2) = P(x>=2)

= 1 - [ 15C0 x (0.04)^0 x (0.96)^15 + 15C1 x (0.04)^1 x (0.96)^14]

= 1 - [0.5420 + 0.3388] .3388]

= 0.1192

d) likelihood that the proportion of the 15 who received a special accommodation is within two standard deviations of the proportion you would anticipate receiving one;

= P( x<=1) = P( x= 0) = 0.5429

e) expected average time = 0.04 x 4.5 + 0.96 x 3 = 3.06hours

Hence, The average time allowed the 15 selected students to be 3.06hours

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