Problem on normal distribution

Gold Member

Homework Statement

The time Rafa spends on his homework each day is normally distributed with mean 1.9hrs and standard deviation σ. On 80% of these days he spends more than 1.35 hours on his homework.
i. find the value of σ
ii. find the probability that Rafa spends less than 2 hours on his homework
iii. A random sample of 200 days is taken. Use an approximation to find the probability that the number of days Rafa spends more than 1.35hrs on homework is between 163days and 173days inclusive

The Attempt at a Solution

i. [/B]i have no problem here
80per cent=0.842
z=-0.842

-0.842=(1.35-1.9)/σ

σ=0.653

ii.

here i have a problem i agree with book that
##Pz=(2-1.9)/0.653##
= P(z is less than 0.1531) now book says that this is equivalent to 0.561 (on one tailed z table) yet i am getting 0.5+0.1531= 0.6531 which is equivalent to 0.394 (on 1 tailed z distribution table).
iii.
text says x→normal distribution with (mean=160,standard deviation=32),
my problem is how did they get σ=32?, i am assuming for 160, they did 80%times 200
the rest of the working that results to
p=(0.9915-0.6707)=0.321 is ok with me.

BvU
Homework Helper
Hi Chwala,

ii) I see the 0.394 (*) appear for a normal distribution with mean 1 and sigma 1 if you ask for the probability that z exceeds 0.5 + 0.1531
However, you want to have the probability that z exceeds 1 + 0.1531 !

 (*) no, that was 0.364 for the probability that z does not exceed 0.5 + 0.1531 in a normal distribution with mean 1 and sigma 1, sorry.

-- How did you find this 0.394 ? I get it when I ask for excel =NORM.DIST(0.731,1,1,TRUE) so the probability that z exceeds 1.269 in a normal distribution with mean 1 and sigma 1​

Last edited:
Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

The time Rafa spends on his homework each day is normally distributed with mean 1.9hrs and standard deviation σ. On 80% of these days he spends more than 1.35 hours on his homework.
i. find the value of σ
ii. find the probability that Rafa spends less than 2 hours on his homework
iii. A random sample of 200 days is taken. Use an approximation to find the probability that the number of days Rafa spends more than 1.35hrs on homework is between 163days and 173days inclusive

The Attempt at a Solution

i. [/B]i have no problem here
80per cent=0.842
z=-0.842

-0.842=(1.35-1.9)/σ

σ=0.653

ii.

here i have a problem i agree with book that
##Pz=(2-1.9)/0.653##
= P(z is less than 0.1531) now book says that this is equivalent to 0.561 (on one tailed z table) yet i am getting 0.5+0.1531= 0.6531 which is equivalent to 0.394 (on 1 tailed z distribution table).
iii.
text says x→normal distribution with (mean=160,standard deviation=32),
my problem is how did they get σ=32?, i am assuming for 160, they did 80%times 200
the rest of the working that results to
p=(0.9915-0.6707)=0.321 is ok with me.

For (ii): if ##Z## is the standard normal (mean = 0, standard deviation = 1) then ##P(\text{hours} \leq 2) = P(Z \leq 0.15302\cdots)##. This will be only slightly greater than 1/2, because the region ##\{ Z \leq 0.153 \}## is not very much larger than the region ##\{ Z \leq 0 \}##. The value ##\text{Answer} = 0.56089## is correct.

You should NOT add 0.1531 to 1/2, because one of these numbers is on the ##z##-axis (that is, the horizontal axis) and the other is on the probability (vertical) axis when you plot the probability density. Adding theses two numbers together is meaningless!

For (iii): on any given day the probability of spending more than 1.35 hours is p = 0.80. In n = 200 (independent!) days, the (random) number N of days where he spends > 1.35 hours is Binomial with parameters n = 200 and p = 0.8. Since n is large, you can approximate this by a normal random variable ##Y## having the same mean and variance as N. What are the mean and variance of a binomial with parameters (n,p) = (200, 0.8)?

BvU
Gold Member
I am still not getting, i wish i could use a schematic diagram here,
in ii) i found 0.1531 being the region between values 2hrs and the mean value μ=1.9 if μ is centrally placed on the normal distribution, then the region satisfying the question as being time spent by Rafa being less than 2 hrs, is the whole region towards the left of the normal distribution. ie the region having probability of 1/2 , on the left hand side of μ=1.9 plus the region 0.1531 or am i my missing something here

Gold Member
for part iii) its clear E(x)= np=##200×0.8=160 ##
Var(x)= npq = ##200×0.8×0.2=32##

Ray Vickson
Homework Helper
Dearly Missed
I am still not getting, i wish i could use a schematic diagram here,
in ii) i found 0.1531 being the region between values 2hrs and the mean value μ=1.9 if μ is centrally placed on the normal distribution, then the region satisfying the question as being time spent by Rafa being less than 2 hrs, is the whole region towards the left of the normal distribution. ie the region having probability of 1/2 , on the left hand side of μ=1.9 plus the region 0.1531 or am i my missing something here

We have
$$P(Z \leq 0.1531) = \underbrace{P(Z \leq 0)}_{=1/2} + \underbrace{P(0 < Z \leq 0.1531)}_{\text{small}}$$
If you want to get the appropriate values from "normal tables" you need to be careful: some tables give ##P(Z \leq z)## for various positive values of ##z##, while others give ##P(0 < Z \leq z)## for various positive values of ##z##. Be sure to check exactly what type of table you are consulting.

Any tables I have seen will tell you exactly what type the are, but a good way to see it for yourself is to look at tabulated values for small ##z > 0##, such as ##z = 0.01## or ##z = 0.1##. If these tabulated probability values are small, that means the table is giving the probability between ##0## and ##z > 0##; if the tabulated values are slightly > 0.5, that means the table is giving the probability to the left of ##z##--that is, from ## -\infty## up to ##z##.

Gold Member
the table am using is having tabulated values slightly greater than 0.5, i.e for z value 0 yields 0.5 and z value 0.1 yields 0.5398, advice

Ray Vickson