MATH 110 Module 4 Exam (2025) / MATH110 Module 4 Exam: Portage Learning, Exams of Nursing

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MATH 110 Module 4 Exam (2025) / MATH110 Module 4

Exam: Portage Learning(Questions & Answers)

MATH 110 Module 4 Exam

Exam Page 1

A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency?

f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) ) n = 8 x = 6, 7, 8 (number of failures) p = 0.

6 failures: n = 8 x = 6 p = 0. n-x = 8-6 = 2 ( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-

7 failures: n = 8 x = 7 p = 0. n-x = 8-7 = 1 ( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-

8 failures: n = 8 x = 8 p = 0. n-x = 8-8 = 0 ( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-

f(6) = 6.21 x 10^-

Exam Page 3

Find each of the following probabilities:

a. Find P(Z ≤ 1.27).

P(Z ≤ 1.27)= .89796.

b. Find P(Z ≥ - .73).

P(Z ≥ - .73)=1- .23270=..

c. Find P(-.09 ≤ Z ≤ .86).

P(-.09 ≤ Z ≤ .86)= .80511- .46414= .34097.

b. Find P(Z ≥ - .73).

1 - p(z ≥ - 0.73) subtract from 1 cause greater than probability

• 0.73 = 0. 1 - 0.23270 = 0.

P(Z ≥ - 0.73) = 0.

c. Find P(-.09 ≤ Z ≤ .86).

p(z ≤ 0.86) - p(z ≤ - 0.09)

• 0.09 = 0. 0.86 = 0.

0.80511 - 0.46414 = 0.

P(-0.09 ≤ Z ≤ 0.86) = 0.

Answer Key

A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 3.7 inches and a standard deviation of .35 inches. If you choose a rod at random, what is the probability that the rod you chose will be:

a) Less than 3.5 inches?

z = (x-u) / o u = 3. o = 0. x = 3.5 (less than) x - u = 3.5 - 3.7 = -0.

-0.2 / 0.35 = -0.5714 = -0. -0.57 on table = 0.

P(Z ≤ -0.57) = 0.

b) Greater than 3.5 inches?

z = (x-u) / o u = 3. o = 0. x = 3.5 (greater than; 1-p(z≤...) x-u = 3.5 - 3.7 = -0.

-0.2/0.35 = -0.5714 = -0.57 = (from table) 0.

1 - 0.28434 = 0.

P(Z ≥ -0.57) = 0.

c) Between 3.4 inches and 4 inches?

Exam Page 4

A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows:

Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data.

Policies Sold Per Day

Probability, f(x)

b) We must find the z-score for x=3.5:

So, we want P(Z ≥-.57). Since this is greater than, we must use: P(Z ≥-.57)=1.0-P(Z ≤-.57)=1-.28434=.

c) Between 3.4 inches and 4 inches?

We must find the z-score for x=3.4:

and the z-score for x=4:

So, we want P(-.86 ≤ Z ≤ .86) P(-.86 ≤ Z ≤ .86)=P(Z≤ .86)-P( Z ≤ - .86). P(-.86 ≤ Z ≤ .86)=.80511-.19489=.61022.

A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows:

Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data.

The expected value is given by E(x)=μ=Σxf(x)=

Policies Sold Per Day

Probability, f(x)

e(x) = u u = ∑xf(x) o^2 = (∑(x-u)^2) x f(x) o = √o

e(x) = (0x0.04) + (1x0.11) + (2x0.23) + (3x0.26) + (4x0.19) + (5x 0.17) = 2.

o^2 = ((0-2.96^2) x 0.4) + ((1-2.96^2)x0.11) + ((2-2.96^2)x0.23) + ((3-2.96^2)x0.26) + ((4- 2.96^2)x0.19) + ((5-2.96^2)x0.17) = 1.

o = √o^ √1.8984 = 1.

Expected number of insurance policies that the salesperson will sell per day is 2. variance = 1. standard dev = 1.

Answer Key