Question

A.) A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.9 years, and standard deviation of 1 years. If you randomly purchase one item, what is the probability it will last longer than 9 years?

B.) A particular fruit's weights are normally distributed, with a mean of 784 grams and a standard deviation of 24 grams. If you pick one fruit at random, what is the probability that it will weigh between 845 grams and 859 grams.

C.) A particular fruit's weights are normally distributed, with a mean of 615 grams and a standard deviation of 11 grams. The heaviest 19% of fruits weigh more than how many grams?
Give your answer to the nearest gram.

D.) A distribution of values is normal with a mean of 228.7 and a standard deviation of 33.7. Find P₈₅, which is the score separating the bottom 85% from the top 15%.
P₈₅ =

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

E.) The combined SAT scores for the students at a local high school are normally distributed with a mean of 1470 and a standard deviation of 303. The local college includes a minimum score of 2137 in its admission requirements.
What percentage of students from this school earn scores that satisfy the admission requirement?
P(X > 2137) = %
Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Part A

Given

Mean = 6.9

SD = 1

We have to find P(X>9)

P(X>9) = 1 – P(X<9)

Z = (X – mean) / SD

Z = (9 – 6.9) / 1 = 2.1

P(Z<2.1) = P(X<9) = 0.982136

P(X>9) = 1 – P(X<9)

P(X>9) = 1 – 0.982136

P(X>9) = 0.017864

Required probability = 0.017864

Part B

Given

Mean = 784

SD = 24

We have to find P(845<X<859)

P(845<X<859) = P(X<859) – P(X<845)

Find P(X<859)

Z = (859 - 784)/24 = 3.125

P(Z<3.125) = P(X<859) = 0.999111

Now find P(X<845)

Z = (845 - 784)/24 = 2.541667

P(Z<2.541667) = P(X<845) = 0.994484

P(845<X<859) = P(X<859) – P(X<845)

P(845<X<859) = 0.999111 - 0.994484

P(845<X<859) = 0.004627

Required probability = 0.004627

Part C

Mean = 615

SD = 11

X = Mean + Z*SD

Z for heaviest 19% = 0.877896

(by using z-table or excel)

X = 615 + 0.877896*11

X = 624.6569

Required Answer = 625 gram

Part D

Mean = 228.7

SD = 33.7

Z for P₈₅ = 1.036433

X = Mean + Z*SD

X = 228.7 + 1.036433*33.7

X = 263.6278

P₈₅ = 263.6278

P₈₅ = 263.6

[***All Z values are taken from z-table or excel.]