z score - the number of standard deviations from the mean a data point is. Also known as a standard score19 increased by twice Gregs score use the variable g to represent Gregs score
19 increased by twice Gregs score use the variable g to represent Gregs score Use g for Greg's score g Twice g means we multiply g by 2: 2g 19 increased by means we add 2g to 19 [B]2g + 19 [MEDIA=youtube]E9a_U7z-fHE[/MEDIA][/B]
A candidate for mayor wants to gauge potential voter reaction to an increase recreational services by estimating the proportion of voter who now use city services. If we assume that 50% of the voters require city recreational services, what is the probability that 40% or fewer voters in a sample of 100 actually will use these city services? First, let's do a test on the proportion using our [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=+40&n=+100&ptype=%3D&p=+0.5&alpha=+0.05&pl=Proportion+Hypothesis+Testing']proportion hypothesis calculator[/URL]: We get Z = -2 Now use the [URL='http://www.mathcelebrity.com/zscore.php?z=p%28z%3C-2%29&pl=Calculate+Probability']z-score calculator[/URL] to get P(z<-2) = [B]0.02275[/B]
A compact disc is designed to last an average of 4 years with a standard deviation of 0.8 years. What is the probability that a CD will last less than 3 years? Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=3&mean=4&stdev=0.8&n=1&pl=P%28X+%3C+Z%29']Z-score and Normal distribution calculator[/URL], we get: [B]0.10565[/B]
A financial analyst computed the ROI for all companies listed on the NYSE. She found that the mean of this distribution was 10% with standard deviation of 5%. She is interested in examining further those companies whose ROI is between 14% and 16% of the approximately 1,500 companies listed on the exchange, how many are of interest of her? First, use our [URL='http://www.mathcelebrity.com/zscore.php?z=p%280.14%3Cz%3C0.16%29&pl=Calculate+Probability']z-score calculator[/URL] to get P(0.14 < z < 0.16) = 0.007889 Divide that by 2 for two-tail test to get0.003944729 Use the NORMSINV(0.003944729) in Excel to get the Z value of 2.66 Therefore, the companies of interest are 2.66 * 1500 * 0.10 = [B]399[/B]
A fuel injection system is designed to last 18 years, with a standard deviation of 1.4 years. What is the probability that a fuel injection system will last less than 15 years? Using our [URL='https://www.mathcelebrity.com/probnormdist.php?xone=15&mean=18&stdev=14&n=1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that: P(X < 15) = [B]0.416834[/B]
A group of students at a school takes a history test. The distribution is normal with a mean of 25, and a standard deviation of 4. (a) Everyone who scores in the top 30% of the distribution gets a certificate. (b) The top 5% of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state? (a) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.70&pl=Calculate+Critical+Z+Value']Top 30% is 70% percentile[/URL] Inverse of normal distribution(0.7) = -0.5244005 Plug into z-score formula, -0.5244005 = (x - 25)/4 [B]x = 22.9024[/B] (b) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']Top 5% is 95% percentile[/URL] Inverse of normal distribution(0.95) = 1.644853627 Plug into z-score formula, 1.644853627 = (x - 25)/4 [B]x = 31.57941451[/B]
A set of 19 scores has a mean of 6.3. A new score of 8 is then included in the data set. What is the new mean? We know the mean formula is: Sum of scores / Number of Scores = Mean We're given mean = 6.3 and number of scores = 19, so we have: Sum of scores / 19 = 6.3 Cross multiply: Sum of scores = 19 * 6.3 Sum of scores = 119.7 Now a new score is added of 8, so we have: Sum of scores = 119.7 + 8 = 127.7 Number of scores = 19 + 1 = 20 So our new mean is: Mean = Sum of scores / Number of Scores Mean = 127.7/20 Mean = [B]6.385[/B] [COLOR=rgb(0, 0, 0)][SIZE=5][FONT=arial][B][/B][/FONT][/SIZE][/COLOR]
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points? For x = 125, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+125&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = 1 P(x < 1) = 0.841345 For x = 85, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+85&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = -1 P(x < -1) = 0.158655 So what we want is the probability between these values:
0.841345 - 0.158655 = [B]0.68269[/B]
Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. a. The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit? b. What proportion of the vehicles would be going less than 50 mph? c. A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion? d. In what way do you think the actual distribution of speeds differs from a normal distribution? a. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=65&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<65) = [B]22.66%[/B] b. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+50&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<50) = [B]0.4269%[/B] c. [URL='http://www.mathcelebrity.com/zcritical.php?a=0.9&pl=Calculate+Critical+Z+Value']Inverse of normal for 90% percentile[/URL] = 1.281551566 Plug into z-score formula: (x - 71)/8 = 1.281551566 [B]x = 81.25241252[/B] d. [B]The shape/ trail differ because the normal distribution is symmetric with relatively more values at the center. Where the actual has a flatter trail and could be expected to occur.[/B]
based on a sample of size 41, a 95% confidence interval for the mean score of all students,, on an aptitude test is from 60 to 66. Find the margin of error
Free Confidence Interval for the Mean Calculator - Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean
Free Critical Z-values Calculator - Given a probability from a normal distribution, this will generate the z-score critical value. Uses the NORMSINV Excel function.
Determine the area under the standard normal curve that lies between: (a) Z = -0.38 and Z = 0.38 (b) Z = -2.66 and Z = 0 (c) Z = -1.04 and Z - 1.67 [B](a) 0.2961 using our [URL='http://www.mathcelebrity.comzscore.php?z=+p%28-0.38%3Cz%3C0.38%29&pl=Calculate+Probability']z score calculator[/URL] (b) 0.4961 using our [URL='http://www.mathcelebrity.com/zscore.php?z=+p%28-2.66%3Cz%3C0%29&pl=Calculate+Probability']z score calculator[/URL] (c) 0.8034 using our [URL='http://www.mathcelebrity.com/zscore.php?z=+p%28-1.04%3Cz%3C1.67%29&pl=Calculate+Probability']z score calculator[/URL][/B]
Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site. On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30. b. Find the 95th percentile, and express it in a sentence. a. P(X >=0.30), calculate the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+0.30&mean=+0.28&stdev=+0.05&n=+1&pl=P%28X+%3E+Z%29']z-score[/URL] which is: Z = 0.4 P(x>0.4) = [B]0.344578 or 34.46%[/B] b. Inverse Normal (0.95) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']calculator[/URL] = 1.644853627 Use NORMSINV(0.95) on Excel 0.28 + 0.05(1.644853627) = [B]0.362242681 or 36.22%[/B]
For a population with ? = 60 and ? = 12, what is the z-score that corresponds to a score of 66? [URL='https://www.mathcelebrity.com/probnormdist.php?xone=66&mean=60&stdev=12&n=1&pl=P%28X+%3C+Z%29']Using our z-score calculator[/URL], we get a probability: [B]0.691462[/B]
Height and weight are two measurements used to track a child's development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean μ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them. a. 11 kg
b. 7.9 kg
c. 12.2 kg a. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+11&mean=10.2&stdev=8&n=+1&pl=1" target="_blank']Answer A[/URL] - Z = 0.1 b. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+7.9&mean=+10.2&stdev=+8&n=+1&pl=1']Answer B[/URL] - Z = -0.288 c. [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+12.2&mean=+10.2&stdev=+8&n=+1&pl=1']Answer C[/URL] - Z = 0.25
Perform a one-sample z-test for a population mean. Be sure to state the hypotheses and the significance level, to compute the value of the test statistic, to obtain the P-value, and to state your conclusion. Five years ago, the average math SAT score for students at one school was 475. A teacher wants to perform a hypothesis test to determine whether the mean math SAT score of students at the school has changed. The mean math SAT score for a random sample of 40 students from this school is 469. Do the data provide sufficient evidence to conclude that the mean math SAT score for students at the school has changed from the previous mean of 475? Perform the appropriate hypothesis test using a significance level of 10%. Assume that ? = 73.
In the wild, monkeys eat an average of 28 bananas a day with a standard deviation of 2 bananas. One monkey eats only 21 bananas. What is the z-score for this monkey? Is the number of bananas the monkey eats unusually low? Using [URL='https://www.mathcelebrity.com/probnormdist.php?xone=21&mean=28&stdev=2&n=1&pl=P%28X+%3C+Z%29']our z-score calculator[/URL], we get: Z < -3.5 P(Z < -3.5) = 0.499767 Also, this [B]is unusually low as it's more than 3 deviations away from the mean[/B]
Kimberly wants to become a member of the desert squad at a big catering company very badly, but she must pass three difficult tests to do so. On the first Terrifying Tiramisu test she scored a 68. On the second the challenging Chocalate-Sprinkled Creme Brulee she scored a 72. If kimberly needs an average of 60 on all three tests to become a member on the squad what is the lowest score she can make on her third and final test This is a missing average problem. Given 2 scores of 68, 72, what should be score number 3 in order to attain an average score of 60? [SIZE=5][B]Setup Average Equation:[/B][/SIZE] Average = (Sum of our 2 numbers + unknown score of [I]x)/[/I]Total Numbers 60 = (68 + 72 + x)/3 [SIZE=5][B]Cross Multiply[/B][/SIZE] 68 + 72 + x = 60 x 3 x + 140 = 180 [SIZE=5][B]Subtract 140 from both sides of the equation to isolate x:[/B][/SIZE] x + 140 - 140 = 180 - 140 x = [B]40[/B]
Free Normal Distribution Calculator - Calculates the probability that a random variable is less than or greater than a value or between 2 values using the Normal Distribution z-score (z value) method (Central Limit Theorem).
Also calculates the Range of values for the 68-95-99.7 rule, or three-sigma rule, or empirical rule. Calculates z score probability
population MEAN OF ENVIRONMENTAL SPECIALIST SALARY IS $62000.A RANDOM SAMPLE OF 45 SPECIALIST IS DRAWN FROM THE POPULATION. WHAT IS THE LIKELIHOOD THAT THE MEAN SALARY SAMPLE IS $59000. ASSUME SIGMA IS $6000. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=59000&mean=62000&stdev=6000&n=45&pl=P%28X+%3C+Z%29']Z-Score calculator[/URL], we get the probability as [B]0.0004[/B].
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based on a sample of size 41, a 95% confidence interval for the mean score of all students,, on an aptitude test is from 60 to 66. Find the margin of error
Free Standard Normal Distribution Calculator - Givena normal distribution z-score critical value, this will generate the probability. Uses the NORMSDIST Excel function.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. a. If X = average distance in feet for 49 fly balls, then X ~ _______(_______,_______)
b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability. Find the probability.
c. Find the 80th percentile of the distribution of the average of 49 fly balls a. N(250, 50/sqrt(49)) = [B]0.42074[/B] b. Calculate Z-score and probability = 0.08 shown [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+240&mean=+250&stdev=+7.14&n=+1&pl=P%28X+%3C+Z%29']here[/URL] c. Inverse of normal distribution(0.8) = 0.8416. Use NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL] Using the Z-score formula, we have 0.8416 = (x - 250)/50 x = [B]292.08[/B]
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. If X = distance in feet for a fly ball, then X ~ b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. a. [B]N(250, 50/sqrt(1))[/B] b. Calculate [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+220&mean=250&stdev=50&n=+1&pl=P%28X+%3C+Z%29']z-score[/URL] Z = -0.6 and P(Z < -0.6) = [B]0.274253[/B] c. Inverse of normal distribution(0.8) = 0.8416 using NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL] Z-score formula: 0.8416 = (x - 250)/50
x = [B]292.08[/B]
Suppose that the manager of the Commerce Bank at Glassboro determines that 40% of all depositors have a multiple accounts at the bank. If you, as a branch manager, select a random sample of 200 depositors, what is the probability that the sample proportion of depositors with multiple accounts is between 35% and 50%? [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=50&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']50% proportion probability[/URL]: z = 2.04124145232 [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=+35&n=+100&ptype==&p=+0.4&alpha=+0.05&pl=Proportion+Hypothesis+Testing']35% proportion probability[/URL]: z = -1.02062072616 Now use the [URL='http://www.mathcelebrity.com/zscore.php?z=p%28-1.02062072616
The average precipitation for the first 7 months of the year is 19.32 inches with a standard deviatiThe average precipitation for the first 7 months of the year is 19.32 inches with a standard deviation of 2.4 inches. Assume that the average precipitation is normally distributed. a. What is the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months? b. What is the average precipitation of 5 randomly selected years for the first 7 months? c. What is the probability of 5 randomly selected years will have an average precipitation greater than 18 inches for the first 7 months? [URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=1&pl=P%28X+%3E+Z%29']For a. we set up our z-score for[/URL]: P(X>18) = 0.7088 b. We assume the average precipitation of 5 [I]randomly[/I] selected years for the first 7 months is the population mean ? = 19.32 c. [URL='https://www.mathcelebrity.com/probnormdist.php?xone=18&mean=19.32&stdev=2.4&n=5&pl=P%28X+%3E+Z%29']P(X > 18 with n = 5)[/URL] = 0.8907
The distribution of actual weights of 8 oz chocolate bars produced by a certain machine is normal with =8.1 ounces and ?=0.1 ounces. A sample of 5 of these chocolate bars is selected. What is the probability that their average weight is less than 8 ounces? Calculate Z score and probability using [URL='http://www.mathcelebrity.com/probnormdist.php?xone=8&mean=8.1&stdev=0.1&n=5&pl=P%28X+%3C+Z%29']our calculator[/URL]: Z = -2.236 P(X < -2.236) = [B]0.012545[/B]
The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What is the probability that a randomly person has an IQ between 85 and 115? b) Find the 90th percentile of the IQ distribution c) If a random sample of 100 people is selected, what is the standard deviation of the sample mean? a) [B]68%[/B] from the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL] b) P(z) = 0.90. so z = 1.28152 using Excel NORMSINV(0.9)
(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]
The weight of a 9.5-inch by 6-inch paperback book published by Leaden Publications, Inc., is 16.2 oz. The standard deviation is 2.9 oz. What is the probability that the average weight of a sample of 33 such books is less than 15.89 oz? Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=15.89&mean=16.2&stdev=2.9&n=33&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we get: [B]0.271[/B]
Free Z Score Lookup Calculator - Given a Z-score probability statement from the list below, this will determine the probability using the normal distribution z-table.
* P(z < a)
* P(z <= a)
* P(z > a)
* P(z >= a)
* P(a < z < b) Calculates z score probability
ncG1vNJzZmivp6x7rq3ToZqepJWXv6rA2GeaqKVfqLKivsKhZamgoHS%2BfsaEa2esm5%2Bnsg%3D%3D