Question

(answer all three parts} Normal (i.e. average) internal temperature for humans is approximately 98 degrees,  If a given population has a standard deviation of 1.2 degrees, what is the maximum temperature for the lowest 35% of the population?  What is the minimum temperature for the highest 25% of the population.

2. If you take a 50 question true-false final exam (two-points each question) and you never paid attention in the class (or knew anything about the topic on your own), what is probability you will receive a grade of 56 or less, in which case you will fail the class?  What is the probability you will get at least 30 questions right, which will  give you a passing grade in the class?

Answer 1

Solution:---

X : Internal temperature of humans

X is normally distributed with mean 98 degrees and standard deviation 1.2 degrees.

Let X35 be the maximum temperature for the lowest 35% of the population i.e

P(X<X35) = 0.35

Let Z35 be the z-score of X35

Z35 = (X35 - mean)/Standard deviation = (X35-98)/1.2

X35 = 98+1.2Z35

P(Z<Z35) = P(X<X35) = 0.35

From standard normal tables ,

P(Z<--0.38) = 0.3520; P(Z<-.39) = 0.3483

Therefore, average of -.38 and -.39 i.e -.385 is a good approximation

P(Z<-.385) = (0.3520+0.3483)/2=0.35015

Z35 = -.385

X35 = 98+1.2Z35 = 98 + 1.2 * (-.385) =98-0.462 = 97.538

Maximum temperature for the lowest 35% of the population = 97.538

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Let X25 minimum temperature for the highest 25% of the population i.e P(X>X25) =0.25

P(X>X25) =1-P(X<X25) = 0.25; P(X<X25) =0.75

Z25 be the Z-score for X25

Z25 = (X25 - mean)/Standard deviation = (X25-98)/1.2

X25 = 98+1.2Z25

P(Z<Z25) = P(X<X25) = 0.75

From standard normla tables,

P(Z<0.67) = 0.7486 ; P(Z<0.68) = 7517

Therefore, average of .67 and .68 i.e .675 is a good approximation

P(Z<.675) = (0.7486+0.7517)/2=0.75015

Z25 = 0.675

X25 = 98+1.2Z25 =98+1.2 * 0.675 = 98+0.81=98.81

minimum temperature for the highest 25% of the population = 98.81

2

This is a binomial distribution question with
n = 50
p = 0.5
q = 1 - p = 0.5
This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know that


The z-score at x = 28.5 is,

z = 0.99
This implies that


The z-score at x = 29.5 is,

z = 1.2728
This implies that