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.

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

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) = ?

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%

26

step-by-step explanation:

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.