**Z-Score Calculation and Analysis****Problem Statement:**A standardized exam's scores are normally distributed. In a recent year, the mean test score was 21.3, and the standard deviation was 5.4. The test scores of four students selected at random are 15, 24, 8, and 36. **Objective:**1. Find the z-scores that correspond to each value. 2. Determine whether any of the values are unusual.**Z-Score Formula:**The z-score is calculated using the formula:\[ z = \frac{(X - \mu)}{\sigma} \]Where:- $ X $ is the individual data point.- $ \mu $ is the mean of the dataset.- $ \sigma $ is the standard deviation.**Task:**Calculate the z-score for the score of 15 (round to two decimal places).**User Interface Options:**- "Help Me Solve This"- "View an Example"- "Get More Help"**Functionality:**- A space to input the z-score calculation.- Options to clear the input and check the answer.

Answered: Image /qna-images/answer/8e2d7ca2-08ea-4dd8-8972-75f1de5388b0.jpg

Answered: A standardized exam's scores are…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Assume that adults have IQ scores that are normally distributed with a mean of 105 and a standard deviation 15 . Find Upper P 3 , which is the IQ score separating the bottom 3 % from the top 97 %. Click to view page 1 of the table. LOADING... Click to view page 2 of the table. LOADING... The IQ score that separates the bottom 3 % from the top 97 % is Upper P 3 equalsnothing.
Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below. Pierce got a score of 71.5; this version has a mean of 62.5 and a standard deviation of 15. Vincent got a score of 309.2; this version has a mean of 294 and a standard deviation of 19. Norma got a score of 7.96; this version has a mean of 7.2 and a standard deviation of 0.4. If the company has only one position to fill and prefers to fill it with the applicant who performed best on the aptitude test, which of applicants should be offered the job?
Use z scores to compare the given values. The tallest living man at one time had a height of 225 cm. The shortest living man at that time had a height of 136.8 cm. Heights of men at that time had a mean of 173.31 cm and a standard deviation of 5.54 cm. Which of these two men had the height that was more extreme?
The weekly mean income of a group of executives is $1,000 and the standard deviation of this group is $100. The distribution is normal. What percent of the executives have an income of $950 or less?
A successful basket ball player has a height of 6 feet 4 inches. Based on statistics his height converts to the z score 2.66. How many standard deviations in his height above the mean
Use z scores to compare the given values. The tallest living man at one time had a height of 246 cm. The shortest living man at that time had a height of 116.2 cm. Heights of men at that time had a mean of 170.86 cm and a standard deviation of 8.41 cm. Which of these two men had the height that was more extreme? and the z score for the shortest man is z = ☐, the man had the height that was more extreme. Since the z score for the tallest man is z = (Round to two decimal places.)
Abby's score of 704 on the math portion of the 2002 GRE exam placed her in the 84th percentile rank. The standard deviation of this exam was 141. If the test scores are normally distributed, what was the mean score of the 2002 test takers? write your answer with one decimal place.
Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Click the icon to view the data table of IQ scores. a. Use a 0.01 significance level to test the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels. What are the null and alternative hypotheses? Assume that population 1 consists of subjects with medium lead levels and population 2 consists of subjects with high lead levels. OA. Ho: H₁ H₂ H₁ H₁ H₂ C. Ho: M₁ = ₂ H₁ H₁ H₂ OB. Ho: H₁ H₂ H₁: µ₁ µ₂ OD. Ho: H#H2 H₁: H₁ H₂
Please answer B, C, & D.
IQ Scores Medium Lead Level O High Lead Level 72 n2 = 11 84 92 X2 = 88.555 85 87 S2 = 10.083 %3D 97 83 92 101 111 91 Print Done
Plz explain. Ty
The mean time taken by all participants to run a road race was found to be 155 minutes with a standard deviation of 15 minutes. Using Chebyshev's theorem, find the percentage of runners who ran this road race in 125 to 185 minutes.Round your answer to the nearest integer.Enter your answer in accordance to the question statement %

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