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ycow [4]
3 years ago
12

Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in r

ural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. Part (a) In words, define the random variable X.
A) the number of 4-year-old children that live in rural areas

B) the time (in hours) a child spends unsupervised per day

C)the number of people that live in rural areas

D)the time (in hours) a 4-year-old spends unsupervised per day

E)the time (in hours) a 4-year-old spends unsupervised per week

Part I: Give the distribution of X.

X ~ ____ (____,_____)

Part II:

Find the probability that the child spends less than 1 hour per day unsupervised.

Write the probability statement.

P_________

What is the probability? (Round your answer to four decimal places.) _____
Mathematics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

D)the time (in hours) a 4-year-old spends unsupervised per day

Step-by-step explanation:

Given that 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe

X - the time (in hours) a 4-year-old spends unsupervised per day

X is N(3,1.8)

P(X

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we have the expression:

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At first glance there is nothing in common, but we can notice that 30 and 70 are multiples of 10, that is:

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2 years ago
Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboa
KengaRu [80]

Answer:

a. <u>0.5 or 50%</u>

b. <u>0.496 or 49.6%</u>

c. <u>0.8185 or 81.85%</u>

d. <u>Yes, it would be just 0.0013 or 0.13% of probability to find a typist whose speed exceeds 105 wpm</u>

e. <u>0.0252 or 2.52%</u>

f. <u>The qualifying speed would be 47.4 wpm.</u>

Step-by-step explanation:

a. Let's find the z-score this way:

μ  = 60 σ= 15

z-score = (x - μ)/σ

z-score = (60 - 60)/15

z-score = 0

Now, let's calculate the p value for z-score = 0, using the z-table:

<u>p (z=0) = 0.5 or 50%</u>

b. z-score = (x - μ)/σ

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Now, let's calculate the p value for z-score = -.0.01, using the z-table:

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If the question were less or equal than 60, a and b would have the same answer. But in this case, the question is "less than 60 wpm".

c.

z-score = (x - μ)/σ

z-score = (45 - 60)/15

z-score = - 1

Now, let's calculate the p value for z-score = -1, using the z-table:

p (z = -1) = 0.1587

z-score = (x - μ)/σ

z-score = (90 - 60)/15

z-score = 2

Now, let's calculate the p value for z-score = 2, using the z-table:

p (z = 2) = 0.9772

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<u>p (-1 ≤ z ≤ 2) = 0.9772 - 0.1587 = 0.8185 or 81.85%</u>

d. z-score = (x - μ)/σ

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Now, let's calculate the p value for z-score = 3, using the z-table:

p (z = 3) = 0.9987

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p = 0.1587 * 0.1587

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z-score = (x - μ)/σ

-0.84 = (x - 60)/15

-12.6 = x - 60

-x = -60 + 12.6

-x = - 47.4

x = 47.4

<u>The qualifying speed would be 47.4 wpm.</u>

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