Math / Data & Statistics

QuestionAccording to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg . Assume that blood pressure is normally distributed.
Round the probabilities to four decimal places. It is possible with rounding for a probability to be 0.0000 . a) State the random variable.
Select an answer b) Find the probability that a randomly selected person in China has a blood pressure of 181.3 mmHg or more. c) Find the probability that a randomly selected person in China has a blood pressure of 185 mmHg or less. $\square$ d) Find the probability that a randomly selected person in China has a blood pressure between 181.3 and 185 mmHg . e) Find the probability that randomly selected person in China has a blood pressure that is at most 70.5 mmHg . f) Is a blood pressure of 70.5 mmHg unusually low for a randomly selected person in China?
Why or why not? $\qquad$ g) What blood pressure do $55 \%$ of all people in China have less than?
Round your answer to two decimal places in the first box. Put the correct units in the second box. $\square$

Studdy Solution

STEP 1

1. Blood pressure is normally distributed with a mean ( $\mu$ ) of 128 mmHg and a standard deviation ( $\sigma$ ) of 23 mmHg.
2. The random variable $X$ represents blood pressure in mmHg.

STEP 2

1. State the random variable.
2. Calculate the probability for blood pressure of 181.3 mmHg or more.
3. Calculate the probability for blood pressure of 185 mmHg or less.
4. Calculate the probability for blood pressure between 181.3 and 185 mmHg.
5. Calculate the probability for blood pressure at most 70.5 mmHg.
6. Determine if 70.5 mmHg is unusually low.
7. Calculate the blood pressure that 55% of people have less than.

STEP 3

The random variable $X$ is defined as the blood pressure of a randomly selected person in China, measured in mmHg.

STEP 4

To find the probability that $X \geq 181.3$ , first calculate the z-score:
$z = \frac{181.3 - 128}{23}$
$z \approx \frac{53.3}{23} \approx 2.3174$
Next, find the probability using the standard normal distribution table or a calculator:
$P(X \geq 181.3) = 1 - P(Z < 2.3174)$
Using a standard normal distribution table or calculator, find $P(Z < 2.3174)$ .
$P(Z < 2.3174) \approx 0.9893$
Thus,
$P(X \geq 181.3) = 1 - 0.9893 = 0.0107$

STEP 5

To find the probability that $X \leq 185$ , calculate the z-score:
$z = \frac{185 - 128}{23}$
$z \approx \frac{57}{23} \approx 2.4783$
Next, find the probability using the standard normal distribution table or a calculator:
$P(X \leq 185) = P(Z < 2.4783)$
Using a standard normal distribution table or calculator, find $P(Z < 2.4783)$ .
$P(Z < 2.4783) \approx 0.9934$

STEP 6

To find the probability that $181.3 \leq X \leq 185$ , use the probabilities calculated in steps 2 and 3:
$P(181.3 \leq X \leq 185) = P(X \leq 185) - P(X < 181.3)$
$P(181.3 \leq X \leq 185) = 0.9934 - 0.9893 = 0.0041$

STEP 7

To find the probability that $X \leq 70.5$ , calculate the z-score:
$z = \frac{70.5 - 128}{23}$
$z \approx \frac{-57.5}{23} \approx -2.5$
Next, find the probability using the standard normal distribution table or a calculator:
$P(X \leq 70.5) = P(Z < -2.5)$
Using a standard normal distribution table or calculator, find $P(Z < -2.5)$ .
$P(Z < -2.5) \approx 0.0062$

STEP 8

A blood pressure of 70.5 mmHg is considered unusually low if the probability is very small. Since $P(X \leq 70.5) \approx 0.0062$ , which is less than 0.05, it is unusually low.

STEP 9

To find the blood pressure that 55% of people have less than, find the z-score corresponding to 0.55 in the standard normal distribution table:
$z \approx 0.1257$
Convert the z-score to the blood pressure value:
$X = \mu + z \cdot \sigma$
$X = 128 + 0.1257 \cdot 23$
$X \approx 128 + 2.8911 \approx 130.89$
The blood pressure is approximately 130.89 mmHg.
The solution to the problem is as follows: a) The random variable $X$ is blood pressure in mmHg. b) $P(X \geq 181.3) = 0.0107$ c) $P(X \leq 185) = 0.9934$ d) $P(181.3 \leq X \leq 185) = 0.0041$ e) $P(X \leq 70.5) = 0.0062$ f) Yes, 70.5 mmHg is unusually low because the probability is less than 0.05. g) 55% of people have a blood pressure less than 130.89 mmHg.

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Studdy Solution

STEP 1

1. Blood pressure is normally distributed with a mean ( $\mu$ ) of 128 mmHg and a standard deviation ( $\sigma$ ) of 23 mmHg.
2. The random variable $X$ represents blood pressure in mmHg.

STEP 2

STEP 3

The random variable $X$ is defined as the blood pressure of a randomly selected person in China, measured in mmHg.

STEP 4

STEP 5

STEP 6

To find the probability that $181.3 \leq X \leq 185$ , use the probabilities calculated in steps 2 and 3:
$P(181.3 \leq X \leq 185) = P(X \leq 185) - P(X < 181.3)$
$P(181.3 \leq X \leq 185) = 0.9934 - 0.9893 = 0.0041$

STEP 7

STEP 8

A blood pressure of 70.5 mmHg is considered unusually low if the probability is very small. Since $P(X \leq 70.5) \approx 0.0062$ , which is less than 0.05, it is unusually low.

STEP 9

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Studdy Solution

STEP 1

1. Blood pressure is normally distributed with a mean (μ\muμ) of 128 mmHg and a standard deviation (σ\sigmaσ) of 23 mmHg.2. The random variable XXX represents blood pressure in mmHg.

STEP 2

STEP 3

The random variable XXX is defined as the blood pressure of a randomly selected person in China, measured in mmHg.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

A blood pressure of 70.5 mmHg is considered unusually low if the probability is very small. Since P(X≤70.5)≈0.0062P(X \leq 70.5) \approx 0.0062P(X≤70.5)≈0.0062, which is less than 0.05, it is unusually low.

STEP 9

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1. Blood pressure is normally distributed with a mean ( $\mu$ ) of 128 mmHg and a standard deviation ( $\sigma$ ) of 23 mmHg.
2. The random variable $X$ represents blood pressure in mmHg.

The random variable $X$ is defined as the blood pressure of a randomly selected person in China, measured in mmHg.

A blood pressure of 70.5 mmHg is considered unusually low if the probability is very small. Since $P(X \leq 70.5) \approx 0.0062$ , which is less than 0.05, it is unusually low.