# Thread: normal distribution help

1. ## normal distribution help

The weights of adult males are distributed N(172,29) (in pounds). This is our shorthand notation for a Normal distribution with a mean of 172 and a standard deviation of 29. Answer the following:

a) what is the probability that an adult male selected at random will weigh between 171 and 182 pounds?

b) what is the probability that if nine men selected at random, their average weight will be between 171 and 182 pounds?

c) if we want to be 90% confident that our sample mean is within two pounds of the actual mean, what is the minimum number of men that must be weight?

d) an adult male is heaver than 15% of other males. what percentage of adult females are lighter than he is? (adult female weights are distributed N(143,29))

2. Originally Posted by inertia
The weights of adult males are distributed N(172,29) (in pounds). This is our shorthand notation for a Normal distribution with a mean of 172 and a standard deviation of 29. Answer the following:

a) what is the probability that an adult male selected at random will weigh between 171 and 182 pounds?

b) what is the probability that if nine men selected at random, their average weight will be between 171 and 182 pounds?

c) if we want to be 90% confident that our sample mean is within two pounds of the actual mean, what is the minimum number of men that must be weight?

d) an adult male is heaver than 15% of other males. what percentage of adult females are lighter than he is? (adult female weights are distributed N(143,29))
a) Pr(171 < X < 182) = Pr(X < 182) - Pr(X < 171). Convert into X = 182 and X = 171 into z-values and use your tables: $\displaystyle Z = \frac{X - \mu}{\sigma}$.

b) $\displaystyle \bar{X}$ ~ Normal $\displaystyle \left(\mu = 172, \sigma = \frac{29}{\sqrt{9}}\right)$. Calculate $\displaystyle \Pr(171 < \bar{X} < 182)$.

c) Go to your formula for the confidence interval of a mean, substitute the appropriate data and solve for n.

d) Find the value of x* such that $\displaystyle \Pr(X < -x^*) = 0.15$. Then calculate Pr(Y < x*) and multiply the answer by 100 to convert into a %.

Note: $\displaystyle \Pr(X < -x^*) = 0.15 \Rightarrow \Pr(X > x^*) = 0.15 \Rightarrow \Pr(X < x^*) = 0.85$.

3. Originally Posted by mr fantastic
a) Pr(171 < X < 182) = Pr(X < 182) - Pr(X < 171). Convert into X = 182 and X = 171 into z-values and use your tables: $\displaystyle Z = \frac{X - \mu}{\sigma}$.

b) $\displaystyle \bar{X}$ ~ Normal $\displaystyle \left(\mu = 172, \sigma = \frac{29}{\sqrt{3}}\right)$. Calculate $\displaystyle \Pr(171 < \bar{X} < 182)$.

c) Go to your formula for the confidence interval of a mean, substitute the appropriate data and solve for n.

d) Find the value of x* such that $\displaystyle \Pr(X < -x^*) = 0.15$. Then calculate Pr(Y < x*) and multiply the answer by 100 to convert into a %.

Note: $\displaystyle \Pr(X < -x^*) = 0.15 \Rightarrow \Pr(X > x^*) = 0.15 \Rightarrow \Pr(X < x^*) = 0.85$.
i get how to do them all accpet part b, can you please explain it a little more? and where does the $\displaystyle \left(\mu = 172, \sigma = \frac{29}{\sqrt{3}}\right)$come from? the sigma = 29/sq.rt. 3?
Thanks

4. Originally Posted by inertia
i get how to do them all accpet part b, can you please explain it a little more? and where does the $\displaystyle \left(\mu = 172, \sigma = \frac{29}{\sqrt{3}}\right)$come from? the sigma = 29/sq.rt. 3?
Thanks
Your notes or textbook should tell you that the sample mean follows a normal distribution with mean $\displaystyle \mu$ and standard deviation $\displaystyle \frac{\sigma}{\sqrt{n}}$ where $\displaystyle \mu$ and $\displaystyle \sigma$ are the population mean and standard deviation respectively.

The actual calculation is done in excatly the same way as in (a).

By the way, I have corrected a typo in my earlier post - it should be $\displaystyle \frac{29}{\sqrt{9}}$, NOT $\displaystyle \frac{29}{\sqrt{3}}$ ....