SOLUTION: Part 1: Binomial Probabilities
A survey said that 3 in 5 students enrolled in higher education took at least one online course last fall. (Remember that we converted the ratio
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A survey said that 3 in 5 students enrolled in higher education took at least one online course last fall. (Remember that we converted the ratio
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Question 1176772: Part 1: Binomial Probabilities
A survey said that 3 in 5 students enrolled in higher education took at least one online course last fall. (Remember that we converted the ratio into a percentage to find the probability that a student took an online class.)
Choose 4 students at random.
a) All 4 students took online courses
b) None of the 4 students took online courses
c) At least 1 of the 4 students took an online course
Include in your discussion what you used for n and p in the binomial distribution and any calculations you used.
Part 2: Use the Standard Normal Table (attached to the end of packet 2).
To qualify for security officers’ training, recruits are tested for stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8. Find the probability that a randomly selected recruit has a score between 64 and 72. Include a detailed description of how you calculated the probability (not just the final answer).
Hi
1) Binomial Distribution: p = .6 and n = 4
a) P(x = 4) = 4C4 (.6)^4(.4)^0 = .1296
0r TI: P(x = 4) = (4,.6,4)
b) P( x = 0) = 4C0 9.60^0(.40)^4 = .0256
c) P( x ≥ 1) = 1 - P(x=0) = 1 - .0256 = .9744
2) Normal distribution: mean of 62 and a standard deviation of 8.
P(64< x < 72) = normalcdf(64,72,62,8) = .2956 |Using TI
that is:
P(x ≤ 72) - P( x≤ 64) 0r P(z ≤ 10/8) - P(z ≤ 2/8)
Using Calculator/ Table to complete either.
Wish You the Best in your Studies.
