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Given that the systolic blood pressure is normally distributed with:

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4 parts
a
0.0146
b
0.2912
c
0.6068
d

79.92 mmHg79.92 \text{ mmHg}

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Part 2

Given that the systolic blood pressure is normally distributed with:
Mean (μ\mu) = 94 mmHg
Variance (σ2\sigma^2) = 121 mmHg
Standard deviation (σ\sigma) = 121=11\sqrt{121} = 11 mmHg

Part a0.0146

What is the probability of a 5-year-old boy having a systolic blood pressure less than 70 mmHg?

  1. Calculate the z-score for x=70x = 70.

    z=x−μσ=70−9411=−2411≈−2.18z = \frac{x - \mu}{\sigma} = \frac{70 - 94}{11} = \frac{-24}{11} \approx -2.18

  2. Find the probability P(X<70)P(X < 70) using the standard normal distribution table.

    P(X<70)=P(Z<−2.18)P(X < 70) = P(Z < -2.18)

    From the Z-table, P(Z<−2.18)≈0.0146P(Z < -2.18) \approx 0.0146.

    The probability is 0.0146\boxed{0.0146}.

Part b0.2912

What is the probability of a 5-year-old boy having a systolic blood pressure higher than 100 mmHg?

  1. Calculate the z-score for x=100x = 100.

    z=x−μσ=100−9411=611≈0.55z = \frac{x - \mu}{\sigma} = \frac{100 - 94}{11} = \frac{6}{11} \approx 0.55

  2. Find the probability P(X>100)P(X > 100) using the standard normal distribution table.

    P(X>100)=P(Z>0.55)=1−P(Z≤0.55)P(X > 100) = P(Z > 0.55) = 1 - P(Z \le 0.55)

    From the Z-table, P(Z≤0.55)≈0.7088P(Z \le 0.55) \approx 0.7088.

    P(Z>0.55)=1−0.7088=0.2912P(Z > 0.55) = 1 - 0.7088 = 0.2912

    The probability is 0.2912\boxed{0.2912}.

Part c0.6068

What is the probability that the blood pressure of a 5-year-old boy will be between 80 and 100 mmHg?

  1. Calculate the z-score for x1=80x_1 = 80.

    z1=80−9411=−1411≈−1.27z_1 = \frac{80 - 94}{11} = \frac{-14}{11} \approx -1.27

  2. Calculate the z-score for x2=100x_2 = 100.

    z2=100−9411=611≈0.55z_2 = \frac{100 - 94}{11} = \frac{6}{11} \approx 0.55

  3. Find the probability P(80<X<100)=P(−1.27<Z<0.55)P(80 < X < 100) = P(-1.27 < Z < 0.55).

    P(−1.27<Z<0.55)=P(Z<0.55)−P(Z<−1.27)P(-1.27 < Z < 0.55) = P(Z < 0.55) - P(Z < -1.27)

    From the Z-table:
    P(Z<0.55)≈0.7088P(Z < 0.55) \approx 0.7088
    P(Z<−1.27)≈0.1020P(Z < -1.27) \approx 0.1020

    P(−1.27<Z<0.55)=0.7088−0.1020=0.6068P(-1.27 < Z < 0.55) = 0.7088 - 0.1020 = 0.6068

    The probability is 0.6068\boxed{0.6068}.

Part d

79.92 mmHg79.92 \text{ mmHg}

What systolic blood pressure is necessary to put someone in the bottom 10% of the distribution?

  1. Find the z-score corresponding to the bottom 10% (i.e., P(Z<z)=0.10P(Z < z) = 0.10).

    From the Z-table, the z-score that corresponds to a cumulative probability of 0.10 is approximately z=−1.28z = -1.28.

  2. Use the z-score formula to solve for xx.

    z=x−μσ  ⟹  x=μ+zσz = \frac{x - \mu}{\sigma} \implies x = \mu + z\sigma x=94+(−1.28)(11)x = 94 + (-1.28)(11) x=94−14.08x = 94 - 14.08 x=79.92x = 79.92

    The systolic blood pressure is approximately 79.92 mmHg\boxed{79.92 \text{ mmHg}}.

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