Unit 3 Saylor MA121 2025/2026 Exam
Questions and Answers | 100% Solved
To find the probability that the mean amount of weight lost by 15 people is
no more than 14 pounds, the random variable should be:
A The number of people who were on the diet.
B The number of people who lost weight on the special weight loss diet.
C The total amount of weight lost by the 15 people on the special weight
loss diet.
D The mean amount of weight lost by the 15 people on the special weight
loss diet. - 🧠 ANSWER ✔✔The mean amount of weight lost by the 15
people on the special weight loss diet.
Find the probability asked for in Question 1. - 🧠 ANSWER ✔✔Let X be the
random variable from Question 1. To find P(X ≤ 14), we use the standard
normal variable Z = (X - 12)/(3/√15) and use a standard normal distribution
table to compute P(Z ≤ (14-12)/(3/√15)) = P(Z ≤ 2.58) = 0.9951.
The answer is: 0.9951
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, Find the 90th percentile for the mean amount of weight lost by 15 people. -
🧠 ANSWER ✔✔From a standard normal distribution table, we obtain P(Z ≤
1.28) = 0.9. Converting back to the random variable X using the relation Z =
(X - 12)/(3/√15) yields X0 = (3/√15)Z0 + 12 = (3/√15)(1.28) + 12 = 12.99.
The answer is 12.99
Assume Ramon has kept track of the times for the first accidents to occur
for 40 different days. Let C = the total cumulative time. Then C follows
which distribution?
A U(0,3)
B Exp (13)
C N(60,5.477)
D N(1.5, 0.01875) - 🧠 ANSWER ✔✔N(60, 5.477)
Using the information in Question 4, find the probability that the total time
for all first accidents to occur is more than 43 hours. - 🧠 ANSWER ✔✔To
compute P(C > 43), we convert to a standard normal variable Z using the
relationship Z = (C - 60)/5.477 and then compute P(Z > (43-60)/5.477) =
P(Z > -3.10) = 0.9990, using a standard normal distribution table.
The answer is: 0.999
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Questions and Answers | 100% Solved
To find the probability that the mean amount of weight lost by 15 people is
no more than 14 pounds, the random variable should be:
A The number of people who were on the diet.
B The number of people who lost weight on the special weight loss diet.
C The total amount of weight lost by the 15 people on the special weight
loss diet.
D The mean amount of weight lost by the 15 people on the special weight
loss diet. - 🧠 ANSWER ✔✔The mean amount of weight lost by the 15
people on the special weight loss diet.
Find the probability asked for in Question 1. - 🧠 ANSWER ✔✔Let X be the
random variable from Question 1. To find P(X ≤ 14), we use the standard
normal variable Z = (X - 12)/(3/√15) and use a standard normal distribution
table to compute P(Z ≤ (14-12)/(3/√15)) = P(Z ≤ 2.58) = 0.9951.
The answer is: 0.9951
COPYRIGHT©NINJANERD 2025/2026. YEAR PUBLISHED 2025. COMPANY REGISTRATION NUMBER:
619652435. TERMS OF USE. PRIVACY STATEMENT. ALL RIGHTS RESERVED 1
, Find the 90th percentile for the mean amount of weight lost by 15 people. -
🧠 ANSWER ✔✔From a standard normal distribution table, we obtain P(Z ≤
1.28) = 0.9. Converting back to the random variable X using the relation Z =
(X - 12)/(3/√15) yields X0 = (3/√15)Z0 + 12 = (3/√15)(1.28) + 12 = 12.99.
The answer is 12.99
Assume Ramon has kept track of the times for the first accidents to occur
for 40 different days. Let C = the total cumulative time. Then C follows
which distribution?
A U(0,3)
B Exp (13)
C N(60,5.477)
D N(1.5, 0.01875) - 🧠 ANSWER ✔✔N(60, 5.477)
Using the information in Question 4, find the probability that the total time
for all first accidents to occur is more than 43 hours. - 🧠 ANSWER ✔✔To
compute P(C > 43), we convert to a standard normal variable Z using the
relationship Z = (C - 60)/5.477 and then compute P(Z > (43-60)/5.477) =
P(Z > -3.10) = 0.9990, using a standard normal distribution table.
The answer is: 0.999
COPYRIGHT©NINJANERD 2025/2026. YEAR PUBLISHED 2025. COMPANY REGISTRATION NUMBER:
619652435. TERMS OF USE. PRIVACY STATEMENT. ALL RIGHTS RESERVED 2
