A random variable X is normally distributed with a mean of 10 and a standard deviation of 3

a) P(X > 12) = 1- P(X ≤ 12) = 1- P(Z = ) = 1- 0.7476 (From normal table) = 0.2524

b) P( X ≤ 5) = P(Z ≤ -5/3) = 0.0478

c) P( 12 ≤ X < 20) = P(X < 20) – P(X < 12) = P(Z < 10/3) – P(Z < 2/3) = 0.9996 - 0.7475 = 0.2521

d) P (X > b) = P(Z >(b-10)/3) = 0.9782 = P(Z >-2.0179)

· b = 10-2.0179*3 = 3.9463

P(A < X < B) = P( = P( -m < Z < m) ; As A and B are symmetric about mean.

= P(Z < m) – P(Z < -m) = P(Z <m ) – 1 +P(Z < m) = 2P(Z < m) – 1.

So,

P(A < X < B) = 0.8592

· 2P(Z < m) – 1 = 0.8592

· P( Z < m) = 0.9296 = P(Z <1.4728)

· m = 1.4728

· B = 95+1.4728*12 = 112.6736

· A = 95-1.4728*12 = 77.3264

Let its weight is k so,

P( X < K) = 0.85

· P(Z < (K-10)/2) = 0.85 = P(Z <1.0364)

· K = 10+2*1.0364 = 12.0729

Thus its weight is 12.0729 pounds.

For simple interest the formula to calculate interest is,

Interest = Principle*Year*Interest rate.

Here,

$85 = $500*year*0.04

· Year = 85/(500*0.04) = 4.25

So the time needed = 4.25 years = 1551.25 days.

We know,

Interest = Principle*Year*Interest rate.

So here,

Principle +Interest = Principle (1 + Year*Interest rate ) = $15250

· Principle =

So the original amount deposited was $14,725.41.

2P = P

Solving this gives,

R = Interest Rate = 3.4808%

Here,

900 = 490

Solving this we have,

T = 12.8251

So we need 12.8251 years for this.

Here,

F = 925= 2191.40

So there will be $2,191.40 in the account after 15 years.

The effective rate is,

Effective rate =

First of all we need to calculate the interest rate,

7500 = 5000

· R = 4.5306%

Now Effective rate = = 0.046082 = 4.6082%

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