Answer to Question #324415 in Statistics and Probability for Gaby

Question

4. Find the value of the number a such that P(0.43 < Z < a) = 0.1359.

5. Suppose that when playing blackjack with a $5 bet, the average (mean) net earning is −$0.25 with a standard deviation of $4.76. You are planning to play blackjack 40 times in a row.

(a) Calculate the probability that your average earning per game will be above $1 per game.

(b) Calculate the probability that you will have suffered a net loss of money (total earning is negative) after playing 40 times.

Answer 1

 4.

P(0.43 < Z < a) = 0.1359.

P(Z<a)-P(Z<0.43)=0.1359

P(Z<0.43)=0.6664

P(Z<a)-0.6664=0.1359

P(Z<a)=0.6664+0.1359

P(Z<a)=0.8023

From the z table an area of 0.8023 correspond to a z value of 0.85

Thus a=0.85

5.

a)

mean($\bar x$ ) =-0.25

standard deviation =4.76

n=40

P(x>1)=?

Z=$\frac{}{}$ $\frac{x-\bar x}{\frac{s}{\sqrt{n}}}$

Z=$\frac{1--0.25}{\frac{4.76}{\sqrt{40}}}$

Z=0.20

P(X>1)=1-P(z$\le$0.2)

P(z$\le$0.2)= 0.5793

P(X>1)=1-0.5793

P(X>1)=0.4207

b)

Probability of a loss is the probability the expected winning is less than 0.

P(x<0)=?

Z=$\frac{0--0.25}{\frac{4.76}{\sqrt{40}}}$

Z=0.04

P(x<0)=P(z<0.04)

P(x<0)=0.5160

The probability of suffering a loss after playing 40 games is 0.5160.