Normal distribution - mean, standard deviation?

Question

I'm stuck on a question from my textbook..

6% of the babies to be born in a hospital have a mass of over 3200g and 14% have a mass of less than 2600g. Given the masses are normally distributed, calculate the mean and the standard deviation.

I'm not just looking for an answer, but an explanation on how you did it please. I found the values of Z but I didn't know how to go about finding the mean and standard deviation as they're both unknowns in the equation.

Thank you! :)

Anonymous · Accepted Answer

p(x>3200) = 0.06 ==> p(x<3200) = 1 - 0.06 = 0.94

p(x<2600) = 0.14

look up these probabilities in a z table and find the corresponding z values

z(0.94) = 1.555

z(0.14) = -1.08

z = (x-m)/s

1.555 = (3200-m)/s

-1.08 = (2600-m)/s

you have 2 equations with 2 unknowns, solve for m and s

2600-m=-1.08s

3200-m = 600+2600-m = 1.555s

600-1.08s = 1.555s

600 = 2.635s

s = 227.7g

m = (2600 + 1.08s) = 2845.92g

? · Answer

P[X > 3200] = 0.06

P[Z > (3200-m)/s] = 0.06

(3200-m)/s = 1.555

3200 - m = 1.555s ---- equation 1

Also,

P[X < 2600] = 0.14

P[Z < (2600-m)/s] = 0.14

(2600-m)/s = -1.09

2600 - m = -1.09s --- equation 2

Solve equations 1&2 for m and s.

Guillermo · Answer

Let X the be mass of the babies to be born in a hospital.

P(X â¥ 3200) = 0.06

P(X â¤ 2600) = 0.14

P((X - Î¼x)/Ïx â¥ (3200 - Î¼x)/Ïx) = 0.06

P(Z â¥ z) = 0.06

z = 1.55

1.55 = (3200 - Î¼x)/Ïx

P((X - Î¼x)/Ïx â¤ (2600 - Î¼x)/Ïx) = 0.14

P(Z â¤ - z) = 0.14

 z = - 1.08

- 1.08 = (2600 - Î¼x)/Ïx
1.55 = (3200 - Î¼x)/Ïx

Ïx = - (2600 - Î¼x)/1.08
Ïx = (3200 - Î¼x)/1.55

- (2600 - Î¼x)/1.08 = (3200 - Î¼x)/1.55

Î¼x = 2,846

Ïx = (3200 - 2846)/1.55

Ïx = 228

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Normal distribution - mean, standard deviation?

3 Answers