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    Normal distribution

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    The World Health Organization has reported that systolic blood pressures of Canadians are normally distributed with a mean of 121 and a standard deviation of 16.
    What symmetric interval about the mean will contain approximately 95% of the systolic blood pressure readings for Canadians?
    A. 105 to 137
    B. 89 to 153
    C. 73 to 169
    D. 89 to 121
    What percentage of Canadians has systolic blood pressure readings greater than 140?
    A. 11.8%
    B. 19.0%
    C. 38.2%
    D. 88.2%

    What systolic blood pressure falls at the 75th percentile?

    A. 91
    B. 110
    C. 132
    D. 196

    What is the probability that a randomly selected Canadian has a systolic blood pressure reading between 110 and 140?

    A. 0.30
    B. 0.50
    C. 0.64
    D. 0.88

    What is the mean and standard deviation for a sampling distribution of 25 Canadians?

    A. Mean = 121 Standard deviation = 16
    B. Mean = 121 Standard deviation = 3.2
    C. Mean = 24.2 Standard deviation = 16
    D. Mean = 24.2 Standard deviation = 3.2

    Correct to 4 decimal places, what is the probability that the mean systolic blood pressure of 25 randomly selected Canadians is greater than 140?

    A. 0.8825
    B. 0.1452
    C. 0.1175
    D. 0.0000

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    https://brainmass.com/statistics/normal-distribution/normal-distribution-181207

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    The World Health Organization has reported that systolic blood pressures of Canadians are normally distributed with a mean of 121 and a standard deviation of 16.

    What symmetric interval about the mean will contain approximately 95% of the systolic blood pressure readings for Canadians?

    A. 105 to 137
    B. 89 to 153
    C. 73 to 169
    D. 89 to 121

    Answer: B. 89 to 153

    95% Confidence limits

    Mean=μ= 121
    Standard deviation =σ= 16
    Confidence level= 95%
    Therefore Significance level=α= 5% =100% -95%
    No of tails= 2
    This is 2 tailed because we are calculating the confidence interval

    Z at the 0.05 level of significance 2 tailed test = 1.96

    Upper confidence limit= μ+z*σ= 152.36 =121+1.96*16
    Lower confidence limit= μ-z*σ= 89.640 =121-1.96*16

    95% Confidence limit: (rounding off the values)
    Upper limit= 153.0
    Lower ...

    Solution Summary

    Answers to multiple choice questions on Normal Distribution, probability calculations for Normal distribution, sampling distribution

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