Question

Computer software generated 300 random numbers that should look as if they are from the standard Normal distribution. They are categorized into five groups: 1) less than or equal to -0.7, 2) greater than 0.7 and less than or equal to -0.3, 3) greater than -0.3 and less than or equal to 0.3, 4) greater than 0.3 and less than or equal to 0.7, and 5) greater than 0.7. The counts in the five groups are 73, 39, 82, 49, and 57, respectively. Compute the expected number for each interval for a sample of 300. Give your answers to two decimal places. 59.80 Expected number for X ≤ -0.7: 60 Expected number for -0.7 < X ≤ -0.3: 30 Expected number for -0.3 < X ≤ 0.3: 60 Expected number for 0.3 < X ≤ 0.7: 60 Expected number for 0.7 < X: 60 Source: More Introduction To The Practice Publisher: W. SouMore ISBN: 10PsbhshenW.

          Computer software generated 300 random numbers that should look as if they are from the standard Normal distribution. They are categorized into five groups: 1) less than or equal to -0.7, 2) greater than 0.7 and less than or equal to -0.3, 3) greater than -0.3 and less than or equal to 0.3, 4) greater than 0.3 and less than or equal to 0.7, and 5) greater than 0.7. The counts in the five groups are 73, 39, 82, 49, and 57, respectively.

Compute the expected number for each interval for a sample of 300. Give your answers to two decimal places.

59.80
Expected number for X ≤ -0.7: 60
Expected number for -0.7 < X ≤ -0.3: 30
Expected number for -0.3 < X ≤ 0.3: 60
Expected number for 0.3 < X ≤ 0.7: 60
Expected number for 0.7 < X: 60

Source: More Introduction To The Practice
Publisher: W. SouMore
ISBN: 10PsbhshenW.

Show more…

computer software generated 300 random numbers that should look as if they are from the standard normal distribution they are categorized into five groups 1 less than or equal to 07 2 greate 23399

Added by Miguel G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

08/11/2022

Step 1

Step 1: Calculate the total count of the 5 groups: 73 + 39 + 82 + 49 + 57 = 300 Show more…

Show all steps

lock

Thanks for your feedback!

Computer software generated 300 random numbers that should look as if they are from the standard Normal distribution. They are categorized into five groups: 1) less than or equal to -0.7, 2) greater than 0.7 and less than or equal to -0.3, 3) greater than -0.3 and less than or equal to 0.3, 4) greater than 0.3 and less than or equal to 0.7, and 5) greater than 0.7. The counts in the five groups are 73, 39, 82, 49, and 57, respectively. Compute the expected number for each interval for a sample of 300. Give your answers to two decimal places. 59.80 Expected number for X ≤ -0.7: 60 Expected number for -0.7 < X ≤ -0.3: 30 Expected number for -0.3 < X ≤ 0.3: 60 Expected number for 0.3 < X ≤ 0.7: 60 Expected number for 0.7 < X: 60 Source: More Introduction To The Practice Publisher: W. SouMore ISBN: 10PsbhshenW.

Play audio

Feedback

Powered by NumerAI

Adi S and 74 other subject Intro Stats / AP Statistics educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 A random number generates 500 numbers that follows normally distributed, standard normal.

00:10 So let us write like this is that follows standard normal distributed.

00:15 And here the data set are given.

00:19 It is seen that less than minus 0 .6, we have 139 and between minus 0 .6 to minus 1 .1, we have 1 .01.

00:28 Between minus 1 .01 to 0 .1, we have 41 counts and between 0 .1 to 0 .6 we have 78 numbers observed, then greater than 0 .6 we have observed 1 .40.

00:41 Now, we have to evaluate the corresponding probabilities, which means that probability of x less than minus 0 .6 is the value.

00:50 And by using the excel formula that is equal to norm.

00:57 D .s.

00:57 Dot dist.

01:01 Since it is standard normal, we can write norm.

01:04 S dot dist, the value minus 0 .6 and then the cumulative values are used true.

01:12 Using this, we have the answer is 0 .2743.

01:17 And for the next probability, that is, the values are between minus minus 6 less than z less than minus 0 .1, then this can be done using probability of z less than minus 0 .1 minus probability of z less than minus 0 .6.

01:39 Since you're subtracting the higher probability minus lower probability, each of this is done using this one itself, that is norm .s.

01:49 Dot s.

01:50 Dot, gist, dot minus 0 .1, true, and minus of this value itself, that is above probability.

02:03 Again, i am writing the equation here also.

02:06 Then, minus 0 .6 true.

02:09 Then, by subtracting these two probabilities, we have the answer, 0 .1859.

02:16 In a similar way, we have evaluated the probability between minus 0 .1 z less than 0 .1 and probability of 0 .1 less than 0 .6, also probability of z greater than 0 .6.

02:37 4 are evaluated and are shown in the table in this picture...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later