Question #323733

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?

1
Expert's answer
2022-04-06T07:22:15-0400

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=xμσ\frac{x-\mu }{\sigma}

sample mean=60

standard deviation =8

1.P(x<52)

z=52608\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=48608\frac{48-60}{8} =-1.5

z2=64608\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\le57)

z=57608\frac{57-60}{8 }

z=-0.375

P(z>57)=1-0.3538

P(z>57)=0.6462

C.

mean =230

s=10

Z=xμσ\frac{x-\mu }{\sigma}

1.

P(x<200)=P(z=20023010)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(z19523010)z\le\frac{195-230}{10})

P(x>195)=1-P(z3.5)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=20023010=3\frac{200-230}{10} =-3

z2=25023010\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.





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