A manufacturing facility produces iron rods whose mean diameter is 2.5 inches with a standard deviation of 0.2 inch. If a sample of 50 metal rods produced by the facility were randomly selected, what is the probability that the mean of these rods will lie between 2.4 inches and 2.8 inches?

According to census statistics, Hispanics comprise 27.5% of the population of New York City. Suppose a random sample of 400 people is selected from New York City. What is the probability that there are fewer than 22% of Hispanics in the sample? Hint: Use Normal approximation to the Binomial distribution.

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  1. Use z-scores.

    First problem:
    z = (x - mean)/(sd/√n)
    Find two z-scores:
    z = (2.4 - 2.5)/(0.2/√50) = ?
    z = (2.8 - 2.5)/(0.2/√50) = ?
    Finish the calculations.
    Next, check a z-table to find the probability between the two scores.

    Find mean and standard deviation.
    mean = np = 400 * .275 = ?
    standard deviation = √npq = √(400 * .275 * .725) = ?
    Note: q = 1 - p
    Finish the calculations.
    Next, use z-scores:
    z = (x - mean)/sd
    z = (.22 - mean)/sd
    Substitute the mean and standard deviation into the z-score formula and calculate, then use a z-table to find the probability.

    I hope this will help get you started.

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  2. Note: For the second problem, we use the normal approximation to the binomial distribution when finding the mean and standard deviation.

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  3. How would you check the Z table for the first problem?

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