We determined the mean and standard deviation of the sampling distribution of x. μ = 70 0 = 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. P(x ≤ a) = P( 2 ≤ ³ = H S o- In other words, we need to find the following. P(x within 0.6 of population mean) = P((70-0.6) ≤ x ≤ (70+ 0.6 ]x) Submit Skip (you cannot come back) = - P(69.4 ≤7 ≤x≤ 70.6 69.4 - 70 1.25 SZS -0.48 Szs -0.48 = =P(z ≤ -0.48 * )- X x - 70 1.25 ]*) P(Z < -0.48) ))
We determined the mean and standard deviation of the sampling distribution of x. μ = 70 0 = 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. P(x ≤ a) = P( 2 ≤ ³ = H S o- In other words, we need to find the following. P(x within 0.6 of population mean) = P((70-0.6) ≤ x ≤ (70+ 0.6 ]x) Submit Skip (you cannot come back) = - P(69.4 ≤7 ≤x≤ 70.6 69.4 - 70 1.25 SZS -0.48 Szs -0.48 = =P(z ≤ -0.48 * )- X x - 70 1.25 ]*) P(Z < -0.48) ))
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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Transcribed Image Text:We determined the mean and standard deviation of the sampling distribution of X.
= 70
My
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.
a
2) = P( ² = ² = 1 =²)
P(x ≤ a)
In other words, we need to find the following.
P(x within 0.6 of population mean) = P((70-0.6) ≤ x ≤ (70+ 0.6
69.4 < X < 70.6
Submit Skip (you cannot come back)
=
=
69.4 - 70
1.25
=
= -0.48 ≤ Z≤ -0.48
P(-0
P(z ≤|
<Z<
<-0.48
X
X
1.25
× ))
X-70
X
- P(Z ≤ -0.48)
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