Every morning there is an 70% chance that I get to the bus on times. if each morning is independent of the others, what is the probability that I get to the bus on times at least 4 times in 5 days? a 0.04437
b. 0.3915
c. 0.05220
d 0.5282
e 0.9657

d 0.5282

Step-by-step explanation:

Given the following :

Probability of getting to the bus on time = 70% = P = probability of success = 0.7

(1 - p) = probability of failure = 0.3

Number of trials(n) = 5 days

Number of sucesses (r) = atleast 4 days : >= 4

This is a binomial probability problem

P(r) = nCr × P^r × (1 - P)^(n - r)

P(X >=4) = P(4) + P(5)

P(4) = 5C4 × (0.7)^4 × (0.3)^1 = 0.36015

P(5) = 5C5 × (0.7)^5 × (0.3)^0 = 0.16807

P(4) + P(5) = 0.36015 + 0.16807 = 0.52822

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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step-by-step explanation:

its the second choice.

step-by-step explanation:

the elements in the range will be 7 less than the elements in the domain.