(b) What is the approximate probability that x will be within 0.6 of the population mean u? We determined the mean and standard deviation of the sampling distribution of x. H-80 -1.25 We also determined that the sampling distribution of x is approximately normal, so we can calculate the desired probability by standardizing. Recall the standardization formula. PUTS 3) = P(25 ----) In other words, we need to find the following. within 0.6 of population mean) - P((80-0.6) sxs (80+ 0.6 - P(79.4 ≤ x ≤ 80.6✔ 79.4-80 1.25 525 80.6✔ 80.6 0.6 )) 80.6 80 1.25
(b) What is the approximate probability that x will be within 0.6 of the population mean u? We determined the mean and standard deviation of the sampling distribution of x. H-80 -1.25 We also determined that the sampling distribution of x is approximately normal, so we can calculate the desired probability by standardizing. Recall the standardization formula. PUTS 3) = P(25 ----) In other words, we need to find the following. within 0.6 of population mean) - P((80-0.6) sxs (80+ 0.6 - P(79.4 ≤ x ≤ 80.6✔ 79.4-80 1.25 525 80.6✔ 80.6 0.6 )) 80.6 80 1.25
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Please help answer STEP 4
![Step 3
(b) What is the approximate probability that will be within 0.6 of the population mean u?
We determined the mean and standard deviation of the sampling distribution of X.
H-80
0 = 1.25
X
We also determined that the sampling distribution of X is approximately normal, so we can calculate the desired probability by standardizing. Recall the standardization formula.
P(x ≤ a) =
)=P(2=²-1=²)
X
In other words, we need to find the following.
P(x within 0.6 of population mean) = P((80 - 0.6) ≤ x ≤ (80 + 0.6✔ 0.6
= P(79.4 ≤ x ≤ 80.6 ✔
80.6
= P(75
Submit
79.480
1.25
Skip (you cannot come back)
≤Z≤
= P(-0..
= Pz≤ 0.48 ✔
=
-0.48 z≤ 0.48
80.6✔
Step 4
Use SALT to find the cumulative probability associated with a particular z-score, rounding the result to four decimal places.
USE SALT
<
P(x within 0.6 of population mean) = P(Z ≤ 0.48) - P(z ≤ -0.48)
= 0.3688
0.3156
x
0.0532
X
80.6 - 80
1.25
0.48)
0.48 ]) - P(Z ≤ -0.48)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1c903f2-e6e2-413b-867e-c476db04ea87%2F246c8dc0-1415-4759-b7f4-a763588b701d%2F1pdwvgse_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 3
(b) What is the approximate probability that will be within 0.6 of the population mean u?
We determined the mean and standard deviation of the sampling distribution of X.
H-80
0 = 1.25
X
We also determined that the sampling distribution of X is approximately normal, so we can calculate the desired probability by standardizing. Recall the standardization formula.
P(x ≤ a) =
)=P(2=²-1=²)
X
In other words, we need to find the following.
P(x within 0.6 of population mean) = P((80 - 0.6) ≤ x ≤ (80 + 0.6✔ 0.6
= P(79.4 ≤ x ≤ 80.6 ✔
80.6
= P(75
Submit
79.480
1.25
Skip (you cannot come back)
≤Z≤
= P(-0..
= Pz≤ 0.48 ✔
=
-0.48 z≤ 0.48
80.6✔
Step 4
Use SALT to find the cumulative probability associated with a particular z-score, rounding the result to four decimal places.
USE SALT
<
P(x within 0.6 of population mean) = P(Z ≤ 0.48) - P(z ≤ -0.48)
= 0.3688
0.3156
x
0.0532
X
80.6 - 80
1.25
0.48)
0.48 ]) - P(Z ≤ -0.48)
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