Normal distribution help with hours of time in a day

Oct 2019
27
1
united states
Suppose 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. 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.

a)What percent of the children spend over 10 hours per day unsupervised? (Round your answer to four decimal places.)

b) 90% of the children spend at least how long per day unsupervised? (Round your answer to two decimal places.)
 

Debsta

MHF Helper
Oct 2009
1,278
582
Brisbane
Same comment as on your other post. Draw a normal distribution marking in mean and interpreting sd.
 
Mar 2012
557
29
you have
mean = 3
standard deviation = 1.8
for a)
required p(t > 10) = ?

you can convert the 10 to Z-table, then you take its result from the table

p(t > 10) = p(z > (10 - mean) / std)

answer = 1 - 0.5 - (table result)
 

Debsta

MHF Helper
Oct 2009
1,278
582
Brisbane
you have
mean = 3
standard deviation = 1.8
for a)
required p(t > 10) = ?

you can convert the 10 to Z-table, then you take its result from the table

p(t > 10) = p(z > (10 - mean) / std)

answer = 1 - 0.5 - (table result)
Depends on the table you are using. So what's your answer?
 
Mar 2012
557
29
My answer is 0.0001

i used Z-table that is cumulative from mean. In other words, area between mean = 0 and z.

z = (10 - 3)/1.8 = 3.89

looking at the table, 3.89 corresponds to area of 0.4999

p(t > 10) = 1 - 0.5 - 0.4999 = 0.0001

you can also get the same result from z-table that is cumulative between negative infinity to z

but here
answer = 1 - (table result)

z = 3.89 corresponds to area of 0.9999

p(t > 10) = 1 - 0.9999 = 0.0001

some tables gives you more digits after the decimal points for more precision but to round your answer to 4 decimal places, they give same answer
 
Mar 2012
557
29
for b)

here the question is tricky but I will assume at least means the required area is 0.90 to the right of the required time

p(t >= hours) = 0.90

my answer is 0.7 hour = 42 minutes

how did I found this answer?
I looked at the z-table for 0.90, using z-table that is accumulative between negative infinity and z,
I found 0.9050 but the number before it is 0.8997 which is more near to 0.90

0.8997 gives z = 1.28

z = (x - mean)/std = 1.28

I need to find x

(x - 3)/1.8 = 1.28
x = 5.304 hours (rounded) = 5.3 hours

but here the area is 90% on the left of the time
5.3 - 3 = 2.3
I am 2.3 hours after the mean

make it before the mean
3 - 2.3 = 0.7