Let X be a random variable representing the mean gray scale of each pixel.
P(110 ≤ X ≤ 140) = P(X < 140) - P(X < 110) = P(z < (140 - 125)/15) - P(z < (110 - 125)/15) = P(z < 1) - P(z < -1) = P(z < 1) - [1 - P(z < 1)] = 2P(z < 1) - 1 = 2(0.84134) - 1 = 1.68268 - 1 = 0.68268
Number of pixels that have gray scale of between 110 and 140 = 0.68268 x 1000000 = 682680 = approximately 680,000
Exact form: 89/3
Decimal form:29.666666.....
Mixed number: 29 2/3
See attachment for math work and answer.
Answer:
a) The Venn diagram is presented in the attached image to this answer.
b) 0.82
c) 0.16
Step-by-step explanation:
a) The Venn diagram is presented in the attached image to this answer.
n(U) = 100%
n(S) = 48%
n(B) = 66%
n(H) = 38%
n(S n B) = 30%
n(B n H) = 22%
n(S n H) = 28%
n(S n B n H) = 12%
The specific breakdowns for each subgroup is calculated on the Venn diagram attached.
b) The probability that a randomly selected student likes basketball or hockey.
P(B U H)
From the Venn diagram,
n(B U H) = n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) = 26 + 10 + 18 + 12 + 16 + 0 = 82%
P(B U H) = 82/100 = 0.82
c) The probability that a randomly selected student does not like any of these sports.
P(S' n B' n H')
n(S' n B' n H') = n(U) - [n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) + n(S n B' n H')]
n(S' n B' n H') = 100 - (26 + 10 + 18 + 12 + 16 + 0 + 2) = 100 - 84 = 16%
P(S' n B' n H') = 16/100 = 0.16