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Let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15. Find to 4 decimal places the probability that x assumes a value a. less than 53 b. greater than 87 c. greater than 54 d. less than 80 eTextbook and Media Save for Later Attempts: 0 of 3 used Submit Answer

Name: let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15 find to 4 decimal places the probability that x assumes a value a less tha 20873
Uploaded: 2024-07-30T13:48:32-08:00
Duration: 1 min 56 s
Channel: Adi S
Description: let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15 find to 4 decimal places the probability that x assumes a value a less tha 20873

          Let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15. Find to 4 decimal
places the probability that x assumes a value
a. less than 53
b. greater than 87
c. greater than 54
d. less than 80
eTextbook and Media
Save for Later
Attempts: 0 of 3 used Submit Answer

let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15 find to 4 decimal places the probability that x assumes a value a less tha 20873

Added by Kathleen V.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Adi S

07/30/2024

Step 1

less than 53 We need to find $P(x < 53)$. First, we calculate the z-score: $z = \frac{x - \mu}{\sigma} = \frac{53 - 72}{15} = \frac{-19}{15} \approx -1.2667$ Using a z-table or calculator, we find the probability: $P(z < -1.2667) \approx 0.1026$ Show more…

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Let x be a continuous random variable that is normally distributed with a mean of 72 and a standard deviation of 15. Find to 4 decimal places the probability that x assumes a value a. less than 53 b. greater than 87 c. greater than 54 d. less than 80 eTextbook and Media Save for Later Attempts: 0 of 3 used Submit Answer