Question

calculate:

PART A: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 203.9-cm and a standard deviation of 0.9-cm. For shipment, 20 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is less than 204.5-cm.
P(M < 204.5-cm) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

PART B: The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.933 g and a standard deviation of 0.327 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 34 cigarettes with a mean nicotine amount of 0.832 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 34 cigarettes with a mean of 0.832 g or less.
P(M < 0.832 g) =  
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

PART C: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 249.1-cm and a standard deviation of 1.6-cm. For shipment, 19 steel rods are bundled together.

Find P₂₂, which is the average length separating the smallest 22% bundles from the largest 78% bundles.
P₂₂ = -cm

Enter your answer as a number accurate to 2 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

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From Z-table, Lookup for Z-value corresponding to area 0.22 to the left of the normal curve.