The average per capita spending on health care in the United Statesis $5274. If the standard deviation is $600 and the distribution of health care spending is approximatelynormal, what is the probability that a randomly selected person spends more than $6000? Find the limits of the middle 50% of individual health careexpenditures.
Find Z:
Z = (x - μ) / σ
Z = (6000 - 5274) / 600
Z = 1.21
Now look at a Z table for normal distributions.
Z = 1.21 corresponds to a probability of 0.8869.
This means that P(X ≤ 6000) = 0.8869.
So P(X > 6000) = 1 - 0.8869 = ______
The middle 50% is bordered by probabilities of 25% (0.25) and 75% (0.75).
From the Z table, these values most closely match Z values of -0.67 and +0.67.
Lower limit:
Z = (x - μ) / σ
-0.67 = (x - 5274) / 600
Solve for x.
Repeat with +0.67 to find x for the upper limit.
Ah, health care spending, the ultimate wallet buster! Alright, let's crunch some numbers and sprinkle some humor on this statistical feast!
To find the probability that a randomly selected person spends more than $6000, we'll need to use the z-score formula. The z-score measures how many standard deviations an individual value is from the mean.
First, let's calculate the z-score for $6000:
z = (x - μ) / σ
Where x is the value we're interested in, μ is the mean, and σ is the standard deviation.
z = ($6000 - $5274) / $600
z = 726 / 600
z ≈ 1.21
Alright, now that we have the z-score, it's time to consult our favorite Z-tables or embrace the magic of technology.
The probability of someone spending more than $6000 can be found by looking up the z-score we calculated (1.21) in the z-table. I know you're dying to know the answer, but let me build up suspense... drumroll, please...
*Clown Bot does an imaginary drumroll using its floppy clown hands*
According to the trusty z-table, the probability of a randomly selected person spending more than $6000 is approximately 11.38%. Yay for probability!
Now, let's move on to the limits of the middle 50% of individual health care expenditures, also known as the interquartile range (IQR). To find those limits, we need to calculate the z-scores for the values that separate the bottom and top 25% of the distribution.
The z-score that corresponds to the bottom 25% is -0.674, and the z-score for the top 25% is 0.674. Why, you ask? Well, because they're the approximate values for a standard normal distribution that I found in my magical statistical encyclopedia. It's like a bearded wizard of knowledge!
Now, using these z-scores, we can calculate the corresponding expenditures, following the same formula as before:
x = z * σ + μ
For the bottom 25%:
x = -0.674 * $600 + $5274
x ≈ $4901.24
And for the top 25%:
x = 0.674 * $600 + $5274
x ≈ $5647.24
These are the limits of the middle 50% of individual health care expenditures, or the IQR. So, based on our calculations, the middle 50% of people spend between approximately $4901.24 and $5647.24 on health care.
And there you have it, the probability of spending more than $6000 and the limits of the middle 50% of health care expenditures, all served with a side of clown humor! I hope I brought some laughter to your stat-filled journey.
To find the probability that a randomly selected person spends more than $6000, we need several pieces of information:
1. Mean (μ): The average per capita spending on health care in the United States is $5274.
2. Standard deviation (σ): The standard deviation of per capita spending on health care is $600.
First, we need to calculate the z-score for $6000 using the z-score formula:
z = (x - μ) / σ
where x is the value we want to find the z-score for.
z = ($6000 - $5274) / $600
z = 726 / $600
z ≈ 1.21
Next, we need to find the probability associated with a z-score of 1.21 using a standard normal distribution table or a calculator. The probability represents the area under the normal curve to the right of the z-score.
Looking up the z-score of 1.21 in a standard normal distribution table, we find that the area to the right is approximately 0.1131. This represents the probability that a randomly selected person spends more than $6000.
Therefore, the probability that a randomly selected person spends more than $6000 is approximately 0.1131 or 11.31%.
To find the limits of the middle 50% of individual health care expenditures, we need to calculate the z-scores associated with the lower and upper quartiles.
The lower quartile (Q1) corresponds to the z-score that represents 25% of the area under the normal curve, and the upper quartile (Q3) corresponds to the z-score that represents 75% of the area under the curve.
Using a standard normal distribution table or a calculator, we find that the z-score for Q1 is approximately -0.6745 and the z-score for Q3 is approximately 0.6745.
To find the corresponding values for health care expenditures, we can use the z-score formula:
x = z * σ + μ
For Q1:
x1 = -0.6745 * $600 + $5274
x1 ≈ $4915.30
For Q3:
x3 = 0.6745 * $600 + $5274
x3 ≈ $5632.70
Therefore, the limits of the middle 50% of individual health care expenditures are approximately $4915.30 and $5632.70.
To solve these questions, we need to use the concepts of Z-score and the standard normal distribution.
To find the probability that a randomly selected person spends more than $6000, we need to calculate the Z-score first.
Z-score is a measure of how many standard deviations a particular value is from the mean. It is calculated using the formula:
Z = (X - μ) / σ,
where X is the value, μ is the mean, and σ is the standard deviation.
In this case, X = $6000, μ = $5274, and σ = $600.
So,
Z = ($6000 - $5274) / $600,
= $732 / $600,
= 1.22.
Now, we need to find the probability of Z being greater than 1.22. We can use a standard normal distribution table or a calculator to find this probability. Looking up the value of 1.22 in the table or using a calculator, we find that the probability is approximately 0.8888. So, the probability that a randomly selected person spends more than $6000 is approximately 0.8888 or 88.88%.
Next, to find the limits of the middle 50% of individual health care expenditures, we need to calculate the Z-scores for the 25th and 75th percentiles.
To find the Z-score for the 25th percentile, we look up the value in the standard normal distribution table or use a calculator. The Z-score for the 25th percentile is approximately -0.674.
To find the Z-score for the 75th percentile, we look up the value in the standard normal distribution table or use a calculator. The Z-score for the 75th percentile is approximately 0.674.
Now, we can calculate the corresponding values of individual health care expenditures using the formula:
X = μ + Z * σ.
For the lower limit, X_lower = $5274 + (-0.674) * $600,
= $5274 - $404.40,
= $4869.60.
For the upper limit, X_upper = $5274 + (0.674) * $600,
= $5274 + $404.40,
= $5678.40.
Therefore, the middle 50% of individual health care expenditures lies between $4869.60 and $5678.40.