Answer:
a. 0.0122
b. 0.294
c. 0.2818
d. 30.671%
e. 2.01 hours
Step-by-step explanation:
Given
Let X represents the number of students that receive special accommodation
P(X) = 4%
P(X) = 0.04
Let S = Sample Size = 30
Let Y be a selected numbers of Sample Size
Y ≈ Bin (30,0.04)
a. The probability that 1 candidate received special accommodation
P(Y = 1) = (30,1)
= (0.04)¹ * (1 - 0.04)^(30 - 1)
= 0.04 * 0.96^29
= 0.012244068467946074580191760542164986632531806368667873050624
P(Y=1) = 0.0122 --- Approximated
b. The probability that at least 1 received a special accommodation is given by:
This means P(Y≥1)
But P(Y=0) + P(Y≥1) = 1
P(Y≥1) = 1 - P(Y=0)
Calculating P(Y=0)
P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)
= 1 * 0.96^36
= 0.293857643230705789924602253011959679180763352848028953214976
= 0.294 --- Approximated
c.
The probability that at least 2 received a special accommodation is given by:
P (Y≥2) = 1 -P(Y=0) - P(Y=1)
= 0.294 - 0.0122
= 0.2818
d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?
First, we calculate the standard deviation
SD = √npq
n = 15
p = 0.04
q = 1 - 0.04 = 0.96
SD = √(15 * 0.04 * 0.96)
SD = 0.758946638440411
SD = 0.759
Mean =np = 15 * 0.04 = 0.6
The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.
P(Y≤2)
P(0) + P(1) + P(2)
=.
P(0) + P(1) = 0.0122 + 0.294
Calculating P(2)
P(2) = (0.04)² * (1 - 0.04)^(30 - 2(
P(2) = 0.00051
So,
P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051
= 0.30671
Thus it 30.671% probable that 0, 1, or 2 students received accommodation.
e.
The expected value from d) is .6
The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours
