A book shelf manufacturer claims that their largest bookshelves can withstand an

Question

A book shelf manufacturer claims that their largest bookshelves can withstand an average of 180 kilograms of books, with a standard deviation of about 27 kilograms. A sample of 43 bookshelves is selected and the weight they could withstand was measured.

a) What is the probability that the value of the sample mean is less than 175.9 or greater than 184.1?

b) What is the probability that the value of the sample mean is less than 178.4?

c) If the sample mean was found to be 178.3, what does this say about the claim? a)The probability that the sample mean would be this low is small, casting doubt on the claimed value of the population mean b)The probability that the sample mean would be this low is not small enough to cast doubt on the claimed value of the population mean c)We do not have enough information to conclude either of these.

Explanation / Answer

Mean ( u ) =180
Standard Deviation ( sd )=27
Number ( n ) = 43
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 175.9) = (175.9-180)/27/ Sqrt ( 43 )
= -4.1/4.1175= -0.9958
= P ( Z <-0.9958) From Standard NOrmal Table
= 0.1597
P(X > 184.1) = (184.1-180)/27/ Sqrt ( 43 )
= 4.1/4.1175 = 0.9958
= P ( Z >0.9958) From Standard Normal Table
= 0.1597
P( X < 175.9 OR X > 184.1) = 0.1597+0.1597 = 0.3194                  
b)
P(X < 178.4) = (178.4-180)/27/ Sqrt ( 43 )
= -1.6/4.1175= -0.3886
= P ( Z <-0.3886) From Standard NOrmal Table
= 0.3488

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