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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)
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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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)
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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