Answer:
The value for k is about k = 11.093kg.
Step-by-step explanation:
To answer this question, that is, P(x<k) = 0.85, we can use the standardized value for a raw score, or the formula for obtaining z-scores:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] [1]
Where
[tex] \\ x[/tex] is the raw score.
[tex] \\ \mu[/tex] is the population mean.
[tex] \\ \sigma[/tex] is the population standard deviation.
In this case, we are asking to find a value x = k, so that 85% of the blocks have weights less than k (kg).
We have also to use the values for the cumulative standard normal distribution with [tex] \\ \mu = 0[/tex] and [tex] \\ \sigma = 1[/tex] to find the value of z that corresponds to a probability of 0.85.
First step: find the z that corresponds to the probability of 0.85.
To use the formula [1], as we previously mentioned, we first need to find the value of z that corresponds to a probability of 0.85 using the cumulative standard normal distribution table (available in any Statistics books or on the Internet). Then, for a cumulative probability of 0.85, the corresponding value for z = 1.03 (approximately).
Second step: solve the formula [1] for x.
Solving this formula for x, we have:
[tex] \\ z = 1.03[/tex]
[tex] \\ \mu = 11kg[/tex]
[tex] \\ \sigma = 0.09kg[/tex]
Then (without units):
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex]
[tex] \\ 1.03 = \frac{x - 11}{0.09}[/tex]
[tex] \\ 1.03*0.09 = x - 11[/tex]
[tex] \\ (1.03*0.09) + 11 = x[/tex]
So
[tex] \\ x = (1.03*0.09) + 11[/tex]
[tex] \\ x = 0.0927 + 11[/tex]
[tex] \\ x = 11.0927 \approx 11.093 = k[/tex]
Thus, this value for x = 11.093 equals the value for k (x = k) so that
[tex] \\ P(x<k) = 0.85[/tex]
Then
[tex] \\ P(x<k=11.093) = 0.8493 \approx 0.85[/tex] or
[tex] \\ P(x<11.093) = 0.8493 \approx 0.85[/tex]
We can see the graph below showing this value of k = 11.093kg for which 85% of the blocks have weights less that it (x = k = 11.093kg).
