# Thread: Statistic practice question help

1. ## Statistic practice question help

The time in minutes that an adult male takes to burn off 5 grams of fat if walking at a steady pace is normally distributed with a mean, μ, equal to 25 minutes and a standard deviation, σ, equal to 5 minutes. Remember to draw diagrams.
a. What proportion of adult males walking at a steady pace take between 26 minutes and 29 minutes to burn off 5 grams of fat?
b. In a random sample of 30 adult males walking at a steady pace what is the probability that the mean time to burn off 5 grams of fat is less than 23 minutes?
c. An adult male is classified as having a slow metabolism if the time taken to burn off 5 grams of fat while walking at a steady pace is in the top 5% of times. What is the cut‐off time for an adult male walking at a steady pace to be classified as having a slow metabolism? Give your answer to the nearest minute.
d. A sport physiologist suggests that 10% of adult males walking at a steady pace can burn off 5 grams of fat in less than 20 minutes. What is the probability that in a random sample of 100 adult males walking at a steady pace less than 8 of them can burn off 5 grams of fat in less than 20 minutes?

2. ## Re: Statistic practice question help

You have a random variable with normal distribution $X\sim {\cal N}(25,5)$.
Its density function is $f_X(t)=\frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(t-\mu)^2}{2 \sigma^2}}=\frac{1}{5 \sqrt{2\pi}} e^{-\frac{(t-25)^2}{50}}$.

Recall that $P(X and $P(a.
And you should know that any normally distributed function can be transformed to have STANDARD normal distribution: $X\sim {\cal N}(\mu,\sigma) \Rightarrow Z=\frac{X-\mu}{\sigma}\sim{\cal N}(0,1)$ meaning that $F_X(x)=F_Z\left(\frac{x-\mu}{\sigma}\right)$ which is great since values of standard normal distribution function are calculated and given in a table so you never have to solve the integrals above.

Here with $X\sim {\cal N}(25,5)$ using transformation as above will give you $Z=\frac{X-25}{5}\sim{\cal N}(0,1)$ and $F_X(x)=F_Z\left(\frac{x-25}{5}\right)$.

a. $P(26

b. $P(X<23)=F_X(23)=F_Z\left(\frac{23-25}{5}\right)=F_Z\left(-\frac{2}{5}\right)=1-F_Z\left(\frac{2}{5}\right)=1-0.6554=0.3446$

c. One has to find the cutoff time T by solving the inequality $P(X. Now, $F_X(T)=0.95$ leads to $F_Z\left(\frac{T-25}{5}\right)=0.95$. Looking at the table you will find that $\frac{T-25}{5}\approx 1.645$. From here you find $T=32.225$ minutes, or rounded to nearest minute $T=32$ minutes.

3. ## Re: Statistic practice question help

As for d:

A sport physiologist suggests that 10% of adult males walking at a steady pace can burn off 5 grams of fat in less than 20 minutes.
That means that for an arbitrarily chosen male there is 10% chance that walking at steady pace he will be able to burn off 5 grams of fat in less than 20 minutes.
What is the probability that in a random sample of 100 adult males walking at a steady pace less than 8 of them can burn off 5 grams of fat in less than 20 minutes?
Here you consider having a random variable with Binomial distribution $X\sim {\cal B}(100,0.1)$. That random variable can be approximated with normally distributed random variable $Y$with distribution parameters $\mu=n\cdot p = 10$ and $\sigma=\sqrt{np(1-p}=3$, $Y\sim {\cal N}(10,3)$.
The answer to the question is
$P(Y<8)=F_Y(8)=F_Z\left(\frac{8-10}{3}\right)=F_Z\left(-\frac{2}{3}\right)=1-F_Z\left(\frac{2}{3}\right)=1-0.7486=0.2514$

4. ## Re: Statistic practice question help

thank you for helping!I got almost all correct except for C and D when I did the question.

5. ## Re: Statistic practice question help

You're welcome mate.