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Question:
On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.
a. In words, define the random variable X
b. What is X ~N(,)
c. Find the probability that the child spends less than 2 hours per day unsupervised.
d. What percent of the children spend over 12 hours per day unsupervised?
Given Information:
Mean = μ = 3.2 hours
Standard deviation = σ = 1.7 hours
Required Information:
a. In words, define the random variable X
b. X ~N(,) = ?
c. P(X < 2) = ?
d. P(X > 12) = ?
Answer:
a) X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b) X ~N(μ,σ) = X ~N(3.2, 1.7)
c) P(X < 2) = 23.88%
d) P(X > 12) = 0%
Explanation:
a)
Let X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b)
X ~N(μ,σ) = X ~N(3.2, 1.7)
Where 3.2 is the average number of hours that 4-year-old child spends being unsupervised and 1.7 is the standard deviation.
c)
We want to find out the probability that a child spends less than 2 hours per day unsupervised.
P(X < 2) = P(Z < (x - μ)/σ)
P(X < 2) = P(Z < (2 - 3.2)/1.7)
P(X < 2) = P(Z < (- 1.2)/1.7)
P(X < 2) = P(Z < -0.71)
The z-score corresponding to -0.71 is 0.2388
P(X < 2) = 0.2388
P(X < 2) = 23.88%
Therefore, the probability that a child spends less than 2 hours per day unsupervised is 23.88%
d)
We want to find out the probability that a child spends over 12 hours per day unsupervised.
P(X > 12) = 1 - P(X < 12 )
P(X > 12) = 1 - P(X < (x - μ)/σ)
P(X > 12) = 1 - P(X < (12 - 3.2)/1.7)
P(X > 12) = 1 - P(X < 8.8/1.7)
P(X > 12) = 1 - P(X < 5.18)
The z-score corresponding to 5.18 is 1
P(X > 12) = 1 - 1
P(X > 12) = 0
Therefore, the probability that a child spends over 12 hours per day unsupervised is 0%
