Answer :
The probability distribution is a) discrete, with b) a mean of 1.23, the standard deviation is c) 0.6458, d) the probability that an American will take 3 or more trips is 0.04, e) the probability that an American will take 2 or less trips is 0.96 and, f) the probability to take at least one trip is 0.94
It is given that the probability distribution of the no. of trips an American adult takes that lasts 5 nights or more
x probability
0 0.06
1 0.70
2 0.20
3 0.03
4 0.01
a)
Here we can see that all the values are isolated and there is no continuity between them. The distribution clearly has breaks between them.
Hence, we can say that the distribution is discrete.
b)
The expected value of any probability
E(X) = ∑xp(x)
where,
x is the random variable
p(x) is the probability of the corresponding random variable hence we get
E(X) = 0 × 0.06 + 1 × 0.70 + 2 × 0.20 + 3 × 0.03 + 4 × 0.01
= 0 + 0.7 + 0.4 + 0.09 + 0.04
= 1.23
Hence the expected value is 1.23
c)
The standard deviation is √σ²
Where σ² = variance of the distribution
σ² = E(X²) - {E(X)}²
{E(X)}² = 1.23²
= 1.5129
E(X²) = ∑x²p(x)
= 0 × 0.06 + 1 × 0.70 + 4 × 0.20 + 9 × 0.03 + 16 × 0.01
= 0.7 + 0.8 + 0.27 + 0.16
= 1.93
Therefore,
σ² = 1.93 - 1.5129
= 0.4171
Therefore
σ = 0.6458
d)
The probability to take 3 or more trips is
P(X≥3)
= P(X = 3) + P(X = 4)
= 0.03 + 0.01
= 0.04
e)
The probability of an American taking 2 or less trips is
1 - P(X≥3)
= 1 - 0.04
= 0.96
f)
The probability that an American will take at least one trip is
P(X≥1)
= 1 - P(X = 0)
= 1 - 0.06
= 0.94
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