Answer to Question #323733 in Statistics and Probability for Mil

Question

ACTIVITY #4

Scoring system: 1 point for correct answer and 2 points for the solution

A. Use the Table of Standard Normal distribution to determine corresponding probabilities of the following:

1. P(Z < -1.45)

2. P(0.5<Z < 2.33)

3. P(Z > 1.78)

B. X is a normally distributed random variable with a mean of 60 and a standard deviation of 8. Find the probabilities indicated by using the table.

1. P(X < 52)

2. P(48 < X < 64)

3. P(X > 57)

C. Solve the following problems by supplying the needed information.

The average number of calories in a 1.8-ounce chocolate bar is 230. Suppose that the distribution of calories is approximately normal with a standard deviation of 10.

1. Find the probability that a randomly selected chocolate bar will have less than 200 calories.

2. Find the probability that a randomly selected chocolate bar will have greater than 195 calories. 3. What is the percentage that a chocolate bar is randomly selected between 200 calories and 250 calories?

Answer 1

Using the standard Z table

A.

1. P(Z<-1.45)=0.0735
2. P(0.5<Z<2.33)=0.9901-0.6915=0.2986
3. P(Z>1.78)=0.0375

B.

Z=$\frac{x-\mu }{\sigma}$

sample mean=60

standard deviation =8

1.P(x<52)

z=$\frac {52-60}{8}$=-1

P(z$\le$-1)= 0.1587

Thus P(x<52) is 0.1587.

2.P(48<x<64)=

P(z₁<z<z₂)

z₁=$\frac{48-60}{8}$ =-1.5

z₂=$\frac{64-60}{8}$ =1.5

P(z₁<z<z₂)=P(z$\le$ 1.5)-P(z$\le$ -1.5)

=0.9332-0.0668

=0.8864

3.P(x>57)=1-P(x$\le$57)

z=$\frac{57-60}{8 }$

z=-0.375

P(z>57)=1-0.3538

P(z>57)=0.6462

C.

mean =230

s=10

Z=$\frac{x-\mu }{\sigma}$

1.

P(x<200)=P($z=\frac{200-230}{10})$

P(z <-3)=0.0014

 The probability that a randomly selected chocolate bar will have less than 200 calories is 0.0014.

2.

P(x >195)=1-P(x<195)

P(x>195)=1-P($z\le\frac{195-230}{10})$

P(x>195)=1-P($z\le -3.5)$

=1-0.0002

=0.9998

 The probability that a randomly selected chocolate bar will more than 195 is 0.9998.

3.

P(200<x<250)=P(z₁<z<z₂)

z₁=$\frac{200-230}{10} =-3$

z₂=$\frac{250-230}{10}$ =2

P(200<x<250)=P(z₂=2)-P(z₁=-3)

=0.9773-0.0014

=0.9760

The percentage that a chocolate bar is randomly selected between 200 calories and 250 calories is 97.60.