Question:
A manufacturer knows that their items have a lengths that are skewed right, with a mean of 11 inches, and standard deviation of 3.2 inches.
If 30 items are chosen at random, what is the probability that their mean length is greater than 12.6 inches?
(Round answer to four decimal places)
Answer:
Given Data:
Given Data:
The population mean (μ)=11
The population standard deviation (σ)=3.2
The sample size (n)=30
→ The probability that the sample mean is greater than 12.6 :
P( x>12.6)=P((x-μ)/(σ/√n)>(12.6-11)/(3.2/√30))=P(z>2.2251)=0.0130 [ use z table or excel ]
→ Final answer = 0.0130
