If I use the distributive property to simplify the expression: 2(6y

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

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Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$