This Question: 1 pt 8 of 21 (20omplete This Test: 21 pts possible Assume that hu

Question

This Question: 1 pt 8 of 21 (20omplete This Test: 21 pts possible Assume that human body temperatures ane normally distributed with a mean of 98.20"F and a standand dvo f0 63" a. Ahospa'ses 100 6·F as the owest temperatro oorsidered to be fever wtat percentage ofnama and he ty pevs uolitecostred to have a fever? Does this percentage suggest that a cutoflf of 100 6"Fis appropriate? b, Physicians want to select a minum temperauro for regum g further medal tests what should himprin betwewrany50%ofheathy people to exceed t?(Such a result is a talse poshe, "hearing t the test resutisposive, but Rnject ¬ realy k) a. The pencentage of normal and healhy persons considered to have a fever is Round to two decimal places as needed) Does this percentage sugest that a outoftf of 100 6"F is appropriate? OA No, because there is a lange probability that a normal and healthy peson would be considered to have ever OB. Yes, because here is a smal probably hat anomal nd healtypem would becond redbha OD. No, becaan there is a smal pratatitythat a norm al and healthy son wold be cons deed to hi Round to two decimal places as needed Click to select your answerls

Explanation / Answer

Persons who have temperature greater than 100.6 is considered to have fever.

P(X>100.6) = P( x-µ/   > (100.6-98.20)/0.62)

P(z > (100.6-98.20)/0.62) = P( z > 3.87) = 1 – p(z<3.87) = 1 - 0.999945582 = 5.44177E-05

Note: to find P(Z <z) use excel function =NORMSDIST(z). Enter the z value to get the probability.

B. Yes because there is a small probability for a a normal and healthy person would be considered to have a fever.

The minimum temperature for requiring for further medical tests for5% of healthy persons:

P(z > x-98.20)/0.62) = 0.95

Note: to find the z value, use excel function =NORMSINV(probability)

z = 1.645

x = z* + µ = 1.645*0.62 + 98.20

x= 99.22 degrees

The minimum temperature is 99.22 degrees.

Probability of a pregnancy lasting more than 308 days:

P( X 300) = P( x-µ/   (308-269)/15)

P( z 2.6) = 1 – p(z <2.6) = 1- 0.9953 = 0.0047 = 0.47%

The z value for the lowest 2% is: -2.054

X = z* + µ = -2.054*15+269

X = 238.19 days

Babies who are born on or below 238 days are considered to be premature.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.