Respuesta :
Answer:
Explained below.
Step-by-step explanation:
Let X = systolic blood pressure measurements.
It is provided that, [tex]X\sim N(\mu=80,\sigma^{2}=3^{2})[/tex].
(a)
Compute the percentage of measurements that are between 71 and 89 as follows:
[tex]P(71<X<89)=P(\frac{71-80}{3}<\frac{X-\mu}{\sigma}<\frac{89-80}{3})[/tex]
            [tex]=P(-3<Z<3)\\=P(Z<3)-P(Z<-3)\\=0.99865-0.00135\\=0.9973[/tex]
The percentage is, 0.9973 × 100 = 99.73%.
Thus, the percentage of measurements that are between 71 and 89 is 99.73%.
(b)
Compute the probability that a person's blood systolic pressure measures more than 89 as follows:
[tex]P(X>89)=P(\frac{X-\mu}{\sigma}>\frac{89-80}{3})[/tex]
        [tex]=P(Z>3)\\=1-P(Z<3)\\=1-0.99865\\=0.00135\\\approx 0.0014[/tex]
Thus, the probability that a person's blood systolic pressure measures more than 89 is 0.0014.
(c)
Compute the probability that a person's blood systolic pressure being at most 75 as follows:
Apply continuity correction:
[tex]P(X\leq 75)=P(X<75-0.5)[/tex]
        [tex]=P(X<74.5)\\\\=P(\frac{X-\mu}{\sigma}<\frac{74.5-80}{3})\\\\=P(Z<-1.83)\\\\=0.03362\\\\\approx 0.034[/tex]
Thus, the probability that a person's blood systolic pressure being at most 75 is 0.034.
(d)
Let x be the blood pressure required.
Then,
P (X < x) = 0.15
⇒ P (Z < z) = 0.15
⇒ z = -1.04
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\\\-1.04=\frac{x-80}{3}\\\\x=80-(1.04\times3)\\\\x=76.88\\\\x\approx 76.9[/tex]
Thus, the 15% of patients are expected to have a blood pressure below 76.9.
(e)
A z-score more than 2 or less than -2 are considered as unusual.
Compute the z score for [tex]\bar x[/tex] as follows:
[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}[/tex]
 [tex]=\frac{84-80}{3/\sqrt{3}}\\\\=2.31[/tex]
The z-score for the mean blood pressure measurement of 3 patients is more than 2.
Thus, it would be unusual.
