A manufacturer claims that its wood treatment protects outdoor wood fences a mea
ID: 3307162 • Letter: A
Question
A manufacturer claims that its wood treatment protects outdoor wood fences a mean of 2,597 days with a standard deviation of 39.7 days. Assume a normal distribution. In a test of the claim, the wood treatment of a randomly selected fence lasted 2540 days.
(a) If the manufacturer’s claim is true, what is the probability that the wood treatment on a randomly selected fence lasts a time less than 2540 days?
(b) Answer the question in (a) if the standard deviation is 24.2 days instead of 39.7 days.
(c) For each of your answers in (a) and (b), how does your answer affect your belief in the manufacturer’s claim? For (a) and for (b) just state your assessment of the manufacturer’s claim, and you will receive full credit for (c)
Explanation / Answer
= 2597 = 39.7 X = 2540
(a) X = + Z
=> 2540 = 2597 + 39.7Z
=> Z = (2540 - 2597) / 39.7 = -1.4358
Probability from Z table is 0.0755.
(b) X = + Z
=> 2540 = 2597 + 24.2Z
=> Z = (2540 - 2597) / 24.2 = -2.3553
Probability from Z table is 0.0093.
(c) The probability in (a) is very low but at an allowance of 0.05 could possibly be accepted.
The probability of (b) is too low to be accepted.
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