[SOLVED]another distribution problem

karush

Well-known member
In a large school, the heights of all $14$yr old students are measured

The heights of the girls are normally distributed with a mean $155$cm and a standard deviation of $10$cm

The heights of the boys are normally distributed with a mean $160$cm and a standard deviation of $12$cm

(a) Find the probability that a girl is taller than $170$cm.

$\frac{155-170}{10}=1.5$

so with $\mu=0$ and $\sigma=1$ then $P(x>1.5) =0.0668072$

View attachment 1090

(b) Given that $10\%$ of the girls are shorter than $x$cm, find $x$

from z-table $10\%$ is about $.25$ so $.25=\frac{x-155}{10} x\approx157$

but i don't think this is the answer $143$ looks closer so ????

there is still (c), (d), and (e) but have to do later

MarkFL

Staff member
a) You have the correct z-score, but I would use:

$$\displaystyle z=\frac{x-\mu}{\sigma}=\frac{170-155}{10}=1.5$$

By my table, the area between 0 and 1.5 is 0.4332, hence:

$$\displaystyle P(X>170)=0.5-0.4332=0.0668$$

b) We want to find the z-score associated with an area of 0.4, which is about 1.28, and we attach a negative sign since this is to the left of the mean.

$$\displaystyle x=z\sigma+\mu=-1.28\cdot10+155=142.2$$

karush

Well-known member
b) We want to find the z-score associated with an area of 0.4, which is about 1.28, and we attach a negative sign since this is to the left of the mean.

$$\displaystyle x=z\sigma+\mu=-1.28\cdot10+155=142.2$$
Where does the "area of 0.4" come from?

(c) Given that $90\%$ of the boys have heights between $q$ cm and $r$ cm

where $q$ and r are symmetrical about $160$ cm, and $q<r$

find the value of $q$ and of $r$.

well I did this half of $90\%$ is $45\%$ so $.45$ on Z table is about $1.66$ so
$160-1.66\cdot12\approx140.3=q$
and
$160+1.66\cdot10\approx179.7=r$

View attachment 1092

is this correct?

Last edited:

MarkFL

Staff member
Where does the "area of 0.4" come from?
We want 90% of the data to be greater than $x$. We know 50% is greater than $\mu$, and so that leaves 40% to be greater than $x$ and less than $\mu$.

MarkFL

Staff member
(c) Given that $90\%$ of the boys have heights between $q$ cm and $r$ cm

where $q$ and r are symmetrical about $160$ cm, and $q<r$

find the value of $q$ and of $r$.

well I did this half of $90\%$ is $45\%$ so $.45$ on Z table is about $1.66$ so
$160-1.66\cdot12\approx140.3=q$
and
$160+1.66\cdot10\approx179.7=r$

View attachment 1092

is this correct?
According to my table, the $z$-score is closer to 1.645 (using linear extrapolation).

Using numeric integration, I find it is closer to 1.64485.

karush

Well-known member
According to my table, the $z$-score is closer to 1.645 (using linear extrapolation).

Using numeric integration, I find it is closer to 1.64485.
0.44950 from the wiki-z-table gave me 1.64

$160-1.64\cdot12\approx140.3=q$
$160+1.64\cdot12\approx179.7=r$
my prev post should of shown 1.64 not 1.66

still have (d) and (e) but have to come back to post it.

karush

Well-known member
In the group of 14yr olds students $60$% are girls and $40$% are boys.

The probability that a girl is taller than $170$cm is $0.066$

The probability that a boy is taller than $170$cm is $0.202$

A fourteen-year-old student is selected at random

(d) Calculate the probability that the student is taller than $170$cm

this is probably not conventional method but if there are $600$ girls then $39$ of them are over $170$ cm and if there are $400$ boys then $81$ of them are over $170$ cm so that is
$\frac{120}{1000}\approx .12$

(e) Given that the student is taller than $170$ cm, what is the probability the student is a girl?
$\frac{39}{120}\approx .33$