Math / Data & Statistics

QuestionHeart Rates For a certain group of individuals, the average heart rate is 71 beats per minute. Assume the variable is normally distributed and the standard deviation is 3 beats per minute. If a subject is selected at random, find the probability that the person has the following heart rate. Use
The Standard Normal Distribution Table. Round the final answers to at least four decimal places and intermediate $z$ -value calculations to two decimal places.
Part: $0 / 3$
Part 1 of 3 (a) Between 67 and 73 beats per minute $P(67<X<73)=$ $\square$

Studdy Solution

STEP 1

1. The heart rate $X$ is normally distributed with a mean $\mu = 71$ beats per minute and a standard deviation $\sigma = 3$ beats per minute.
2. We need to find the probability that the heart rate is between 67 and 73 beats per minute.

STEP 2

1. Convert the heart rates to $z$ -scores.
2. Use the Standard Normal Distribution Table to find the probabilities.
3. Calculate the probability for the range.

STEP 3

Convert the lower bound (67 beats per minute) to a $z$ -score using the formula:
$z = \frac{X - \mu}{\sigma}$
For $X = 67$ :
$z = \frac{67 - 71}{3} = \frac{-4}{3} \approx -1.33$

STEP 4

Convert the upper bound (73 beats per minute) to a $z$ -score using the same formula:
For $X = 73$ :
$z = \frac{73 - 71}{3} = \frac{2}{3} \approx 0.67$

STEP 5

Use the Standard Normal Distribution Table to find the probability corresponding to the $z$ -scores.
For $z = -1.33$ , find $P(Z < -1.33)$ .
For $z = 0.67$ , find $P(Z < 0.67)$ .

STEP 6

From the Standard Normal Distribution Table:
$P(Z < -1.33) \approx 0.0918$
$P(Z < 0.67) \approx 0.7486$

STEP 7

Calculate the probability that the heart rate is between 67 and 73 beats per minute:
$P(67 < X < 73) = P(Z < 0.67) - P(Z < -1.33)$
$P(67 < X < 73) = 0.7486 - 0.0918 = 0.6568$
The probability that the heart rate is between 67 and 73 beats per minute is:
$\boxed{0.6568}$

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Studdy Solution

STEP 1

1. The heart rate $X$ is normally distributed with a mean $\mu = 71$ beats per minute and a standard deviation $\sigma = 3$ beats per minute.
2. We need to find the probability that the heart rate is between 67 and 73 beats per minute.

STEP 2

1. Convert the heart rates to $z$ -scores.
2. Use the Standard Normal Distribution Table to find the probabilities.
3. Calculate the probability for the range.

STEP 3

Convert the lower bound (67 beats per minute) to a $z$ -score using the formula:
$z = \frac{X - \mu}{\sigma}$
For $X = 67$ :
$z = \frac{67 - 71}{3} = \frac{-4}{3} \approx -1.33$

STEP 4

Convert the upper bound (73 beats per minute) to a $z$ -score using the same formula:
For $X = 73$ :
$z = \frac{73 - 71}{3} = \frac{2}{3} \approx 0.67$

STEP 5

Use the Standard Normal Distribution Table to find the probability corresponding to the $z$ -scores.
For $z = -1.33$ , find $P(Z < -1.33)$ .
For $z = 0.67$ , find $P(Z < 0.67)$ .

STEP 6

From the Standard Normal Distribution Table:
$P(Z < -1.33) \approx 0.0918$
$P(Z < 0.67) \approx 0.7486$

STEP 7

Calculate the probability that the heart rate is between 67 and 73 beats per minute:
$P(67 < X < 73) = P(Z < 0.67) - P(Z < -1.33)$
$P(67 < X < 73) = 0.7486 - 0.0918 = 0.6568$
The probability that the heart rate is between 67 and 73 beats per minute is:
$\boxed{0.6568}$

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Studdy Solution

STEP 1

1. The heart rate XXX is normally distributed with a mean μ=71\mu = 71μ=71 beats per minute and a standard deviation σ=3\sigma = 3σ=3 beats per minute.2. We need to find the probability that the heart rate is between 67 and 73 beats per minute.

STEP 2

1. Convert the heart rates to zzz-scores.2. Use the Standard Normal Distribution Table to find the probabilities.3. Calculate the probability for the range.

STEP 3

Convert the lower bound (67 beats per minute) to a zzz-score using the formula:z=X−μσ z = \frac{X - \mu}{\sigma} z=σX−μ​For X=67X = 67X=67:z=67−713=−43≈−1.33 z = \frac{67 - 71}{3} = \frac{-4}{3} \approx -1.33 z=367−71​=3−4​≈−1.33

STEP 4

Convert the upper bound (73 beats per minute) to a zzz-score using the same formula:For X=73X = 73X=73:z=73−713=23≈0.67 z = \frac{73 - 71}{3} = \frac{2}{3} \approx 0.67 z=373−71​=32​≈0.67

STEP 5

Use the Standard Normal Distribution Table to find the probability corresponding to the zzz-scores.For z=−1.33z = -1.33z=−1.33, find P(Z<−1.33)P(Z < -1.33)P(Z<−1.33).For z=0.67z = 0.67z=0.67, find P(Z<0.67)P(Z < 0.67)P(Z<0.67).

STEP 6

From the Standard Normal Distribution Table:P(Z<−1.33)≈0.0918 P(Z < -1.33) \approx 0.0918 P(Z<−1.33)≈0.0918P(Z<0.67)≈0.7486 P(Z < 0.67) \approx 0.7486 P(Z<0.67)≈0.7486

STEP 7

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1. The heart rate $X$ is normally distributed with a mean $\mu = 71$ beats per minute and a standard deviation $\sigma = 3$ beats per minute.
2. We need to find the probability that the heart rate is between 67 and 73 beats per minute.

1. Convert the heart rates to $z$ -scores.
2. Use the Standard Normal Distribution Table to find the probabilities.
3. Calculate the probability for the range.

Convert the lower bound (67 beats per minute) to a $z$ -score using the formula:
$z = \frac{X - \mu}{\sigma}$
For $X = 67$ :
$z = \frac{67 - 71}{3} = \frac{-4}{3} \approx -1.33$

Convert the upper bound (73 beats per minute) to a $z$ -score using the same formula:
For $X = 73$ :
$z = \frac{73 - 71}{3} = \frac{2}{3} \approx 0.67$

Use the Standard Normal Distribution Table to find the probability corresponding to the $z$ -scores.
For $z = -1.33$ , find $P(Z < -1.33)$ .
For $z = 0.67$ , find $P(Z < 0.67)$ .

From the Standard Normal Distribution Table:
$P(Z < -1.33) \approx 0.0918$
$P(Z < 0.67) \approx 0.7486$