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Problem 33

# Suppose that diameters of a shaft s manufactured by a certain machine are normal random variables with mean 10 and s.d. $$0.1 .$$ If for a given application the shaft must meet the requirement that its diameter falls between $$9.9$$ and $$10.2 \mathrm{~cm}$$. What proportion of shafts made by this machine will meet the requirement?

### Short Answer

Expert verified
The proportion of shafts produced by this machine that will meet the requirement is 0.8185
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## Step 1: Define the problem in terms of Z-scores

The problem can be defined in terms of Z-scores. The Z-score of an observation is given by $$Z = \frac{x- \mu}{\sigma}$$, where $$x$$ is the observation, $$\mu$$ is the mean and $$\sigma$$ is the standard deviation. In our case, the two Z-scores for the interval [9.9 cm, 10.2 cm] would be calculated as, $$Z_{1} = \frac{9.9 - 10}{0.1} = -1$$, $$Z_{2} = \frac{10.2 - 10}{0.1} = 2$$.

## Step 2: Look up the corresponding areas in the standard normal table

The Z-scores give us the number of standard deviations away from the mean. Using the standard normal table, we can find the area to the left of the Z-score which represents the cumulative probability. For $$Z=-1$$, the area is 0.1587 which represents the probability that the diameter is less than 9.9 cm. The area for $$Z=2$$ is 0.9772, which represents the probability that the diameter is less than 10.2 cm.

## Step 3: Subtract the two areas to get the proportion

The probability that the diameter is between 9.9 cm and 10.2 cm is the difference of the two areas. So, the required proportion, \(P(9.9 < X < 10.2) = P(X < 10.2) - P(X < 9.9) = 0.9772 - 0.1587 = 0.8185.

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