Answer:
a. 0.0122
b. 0.294
c. 0.2818
d. 30.671%
e. 2.01 hours
Step-by-step explanation:
Let X denote the number of students who get special accommodation.
P(X) = 4%
P(X) = 0.04
Let S represent the sample size, which equals 30.
Let Y indicate the chosen numbers within the sample size.
Y follows a Binomial distribution, Bin(30,0.04).
a. The chance that exactly 1 student 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 — rounded
b. The probability that at least 1 received special accommodation is calculated as:
This implies P(Y≥1).
Since P(Y=0) + P(Y≥1) = 1, we find that P(Y≥1) = 1 - P(Y=0).
Now calculating P(Y=0):
P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)
= 1 * 0.96^30
= 0.293857643230705789924602253011959679180763352848028953214976
= 0.294 — rounded
c.
Determining the probability that at least 2 students received special accommodation:
P(Y≥2) = 1 - P(Y=0) - P(Y=1)
= 0.294 - 0.0122
= 0.2818
d. To find the probability that among the 15 students, the received special accommodation” falls within 2 standard deviations from what we anticipate,
we first compute 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
Average = np = 15 * 0.04 = 0.6.
The range that embodies two standard deviations away from 0.6 is [0, 2.55], meaning we are interested in the probability that 0, 1, or 2 students out of 20 received special accommodation.
P(Y≤2)
P(0) + P(1) + P(2)
=.
P(0) + P(1) = 0.0122 + 0.294.
To find P(2):
P(2) = (0.04)² * (1 - 0.04)^(30 - 2)
P(2) = 0.00051.
Thus,
P(0) + P(1) + P(2) = 0.0122 + 0.294 + 0.00051
= 0.30671.
Therefore, it’s 30.671% likely that 0, 1, or 2 students received accommodations.
e.
The anticipated value from part d) is 0.6.
The expected time is calculated as [.6(4.5) + 19.2(3)]/30 = 2.01 hours.
