Finding standard deviation and mean

**loadup2** · September 30th 2009, 03:22 PM

I draw a standard normal curve and filled out everything but I couldn't figure it out. Can someone give me some hints?

ok. so mean and standard deviation are unknown. This guy spends less than one minute doing something about 40% of time and spends more than two minutes doing something 2% of time.

Any help would be appreciated.

**mr fantastic** · September 30th 2009, 11:00 PM

Originally Posted by loadup2

I draw a standard normal curve and filled out everything but I couldn't figure it out. Can someone give me some hints?

ok. so mean and standard deviation are unknown. This guy spends less than one minute doing something about 40% of time and spends more than two minutes doing something 2% of time.

Any help would be appreciated.

$\Pr(X < 1) = 0.4$ :

$\Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533$ . Therefore:

$Z = \frac{X - \mu}{\sigma} \Rightarrow -0.2533 = \frac{1 - \mu}{\sigma}$ .... (1)

$\Pr(X > 2) = 0.02$ :

$\Pr(Z > z_2) = 0.02 \Rightarrow z_2 = 2.054$ . Therefore:

$Z = \frac{X - \mu}{\sigma} \Rightarrow 2.054 = \frac{2 - \mu}{\sigma}$ .... (2)

Solve equations (1) and (2) simultaneously.

**kburk46** · November 23rd 2009, 02:15 PM

Why is $\Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533$ ?

**mr fantastic** · November 23rd 2009, 04:00 PM

Originally Posted by kburk46

Why is $\Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533$ ?

Because Pr(Z < -0.2533) = 0.4.

**kburk46** · November 23rd 2009, 06:31 PM

but how did you come up with -.2533?

**mr fantastic** · November 23rd 2009, 06:36 PM

Originally Posted by kburk46

but how did you come up with -.2533?

Pr(Z < z1) = 0.4 => Pr(Z > -z1) = 0.4 => Pr(Z < -z1) = 0.6 at which point you go to your tables and get the value of -z1 and hence z1.

Since z1 is known to be negative before any calculation is done (why?) it might be easier to see it as:

Pr(Z < -a) = 0.4 => Pr(Z > a) = 0.4 => Pr(Z < a) = 0.6 and now go to tables. Then a = 0.2533 => -a = -0.2533.

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