4. Solve the system using substitution.
1 answer:
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Answer:
12.1%
Step-by-step explanation:
Given that:
Mean (μ) = 20.2 grams and standard deviation (σ) = 0.18 grams.
The z score is a score used to determine the number of standard deviations by which the raw score is above or below the mean. A positive z score means that the raw score is above the mean and a negative z score means that the raw score is below the mean. It is given by:
a) For x < 19.99 g:
From the normal distribution table, P(x < 19.99) = P(z < -1.17) = 0.1210 = 12.1%
The probability that a randomly chosen mouse has a mass of less than 19.99 grams is 12.1%
Complete question:
Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.
1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)
Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.
1b. Find the probabilities of each of these three events. A2, A5, A4
1c. Do the events A2, A5, A4 constitute a partition of the sample space?
Answer:
Given:
n = 20
a) The possible outcome in the sample space,S, =
ⁿCₓ = ²⁰C₅
= 15504
b) probabilities of A2, A5, A4
P(A2) = ⁵C₂ / ²⁰C₂
P(A5) = ⁸C₅ / ²⁰C₅
P(A4) = ⁷C₄ / ²⁰C₄
c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1