3 assume standard normal distribution a find pz 1645 text or...

Name: 3 assume standard normal distribution a find pz 1645 text or z 175 10 points drawing and shading required b find k such that pz k 0025 drawing and shading required 10 points 4 the time spent 41033
Uploaded: 2022-01-16T17:14:31-08:00
Duration: 8 min 41 s
Channel: Robin Corrigan
Description: 3 assume standard normal distribution a find pz 1645 text or z 175 10 points drawing and shading required b find k such that pz k 0025 drawing and shading required 10 points 4 the time spent 41033

#3 Assume Standard normal distribution. (a) Find $P(Z < -1.645 \text{ or } Z > 1.75)$ (10 points) Drawing and shading Required. (b) Find k such that $P(Z > k) = 0.025$. Drawing and shading Required. (10 points) #4 The time spent studying for an exam and the result of the exam for a random sample of 10 students in a math course are shown. (20 points) Time in hours 5 7 5 8 7 10 5 8 9 8 Exam's score 65 80 75 78 75 95 85 90 95 75 Predict the exam result for someone who studied about 4 hours assuming the linear correlation is significant. Round your answer to a whole number.

Transcript

00:01 In this exercise, we were told that the time that it takes to complete exam.

00:06 We'll call that random variable x is normally distributed with the mean of 75 and standard deviation of 10.

00:17 And then we're asked in part a to find the probability that a randomly selected student will complete the exam in more than 80 minutes.

00:27 So that is the probability that x is greater than 80.

00:35 We can re -express this as 1 minus the probability that x is at most 80.

00:45 And we can solve this probability in excel.

00:48 So we enter our formula in excel, we start with equals.

00:52 We use the normal distribution function.

00:54 So that's the first one highlighted in blue.

00:58 We want the probability that x is lessen or equal to 80.

01:02 So we enter 80.

01:04 And then we enter the parameters for our distribution.

01:06 The mean is 75.

01:08 Standard deviation is 10.

01:10 And for cumulative we want true because we are looking for a cumulative probability.

01:14 The probability that x is less than or equal to 80.

01:19 And we get 0 .6915.

01:34 And so that probability comes out to 0 .3085.

01:40 Then for b, we want to find the probability that among 100 students who take the exam, at least one will complete it in less than 65 minutes.

01:50 For step one, let's find the probability that a randomly selected student completes the exam in less than 65 minutes.

02:06 And we'll do this in excel.

02:07 You can recycle the formula we just used, except we enter 65 instead of 80 for the first argument, and it's .1587.

02:26 And let's refer to this probability as p.

02:30 Now we have 100 students in our sample, and let's define a random variable y as the number of students in the sample of 100 who complete the exam in less than 65 minutes.

03:01 Here, why is a binomial random variable? because we have bernoulli trials, 100 of them.

03:07 So a bernoulli trial is where you only have two possibilities, success and failure, and each trial is independent from the other trials.

03:17 And we know they're independent because the distribution for each student is that normal distribution of mean 75 minutes and standard deviation of 10 minutes.

03:27 Now for a binomial random variable, the probability mass function is given by this formula.

03:56 It's n choose x times p to the exponent x times 1 minus p to the exponent n minus x.

04:04 And x can be any integer from 0 up to n...

Question

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