To qualify for a medical study, an applicant must have systolic blood pressure in the 50% of the middle range. If the systolic blood pressure is normally distributed with a mean of 120 and a standard deviation of 4, find the upper and lower limits of the blood pressure a person must have to qualify for the study.

Respuesta :

Answer:

Lower limit: 117.3

Upper limit: 122.7

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 120, \sigma = 4[/tex]

Middle 50%

So it goes from X when Z has a pvalue of 0.5 - 0.5/2 = 0.25 to X when Z has a pvalue of 0.5 + 0.5/2 = 0.75

Lower limit

X when Z has a pvalue of 0.25. So X when [tex]Z = -0.675[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 120}{4}[/tex]

[tex]X - 120 = -0.675*4[/tex]

[tex]X = 117.3[/tex]

Upper limit

X when Z has a pvalue of 0.75. So X when [tex]Z = 0.675[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 120}{4}[/tex]

[tex]X - 120 = 0.675*4[/tex]

[tex]X = 122.7[/tex]

The upper and lower limits of the blood pressure a person must have to qualify for the study are;

Lower limit: 117.3

Upper limit: 122.7

We are given;

Mean; μ = 120

Standard deviation; σ = 4

Pressure must be in the middle range which is 50%.

This means that;

Lower range = 25% = 0.25

Upper range = 75% = 0.75

This means;

Upper limit will be the sample mean at the z-value of the p-value of 0.25

Lower limit will be the sample mean at the z-value of the p-value of 0.25

From online z-score from p-value calculator, we have;

z at p-value of 0.25; z = -0.6745

z at p-value of 0.75; z = 0.6745

Formula for z-score is;

z = (x' - μ)/σ

Making x' the subject;

x' = zσ + μ

Thus;

Upper limit = (0.6745 * 4) + 120

Upper limit ≈ 122.7

Lower limit = (-0.6745 * 4) + 120

Lower limit ≈ 117.3

Read more about normal distribution at; https://brainly.com/question/25800303