# Normal Distribution

**Topics:**Normal distribution, Standard deviation, Standard error

**Pages:**3 (809 words)

**Published:**November 14, 2012

Ans : a) P( Z > 2.58) = 0.0049 ( 4 decimal places)

b) P ( Z < -1) = 0.1587 ( 4 decimal places)

c) P ( -1.5≦ Z < 5) = P ( -1.5 < Z < 5)

= (0.5- 0.0668) + ( 0.5 -0) = 0.9332 ( 4 decimal places)

2.Find the value of z if the area under a Standard Normal curve a)to the right of z is 0.3632;

b)to the left of z is 0.1131;

c)between 0 and z, with z > 0, is 0.4838;

d)between -z and z, with z > 0, is 0.9500.

Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table) b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table) c ) the area between 0 to z is 0.4838, z = 2.14

d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96

3.Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find a)P(X < 15);

b)the value of k such that P(X < k) = 0.2236;

c)the value of k such that P(X > k) = 0.1814;

d)P( 17 < X < 21).

Ans : X ~ N ( 18, 2.52)

a) P ( X < 15)

P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)

b) P ( X < k) = 0.2236

P ( Z < ( k – 18) / 2.5 ) = 0.2236

From normal table, 0.2236 = -0.76

(k-18)/2.5 = - 0.76, solve k = 16.1

c) P (X > k) = 0.1814

P ( Z > (k-18)/2.5 ) = 0.1814

From normal table, 0.1814 = 0.91

(k-18)/ 2.5 = 0.91, solve k = 20.275

d) P ( 17 < X < 21)

P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)

P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)

4.In a sample of 25 observations from a Normal Distribution with mean 98.6 and standard deviation 17.2, find:

Ans: a) n = 25, [pic] = ( = 98.6, [pic] = /n = 17.2/(25 = 3.44 [pic]( N...

Please join StudyMode to read the full document