cm74973551..65x7465189*6.71-38lsbeen 159.68 cm+447495.1*2.5 (9)TEIC. P(Z < 2.13)$0.0 (9)and 178.7 cmd. P(-0.45 < z < 2.29)J4=1447 XSeight of 15 to 18-old males from Chile2= X-Me. Find c if P(Z > c) = 0.39.30.0 (b)PEA.88 (b)521f. Find c if P(-c 1.24) 29tunima.II (b)Page 1 of 4SX2X=4+2(0)SmolברךVNIHO NI LName:zwod 3E.IS equor 8 (5)88.0 (p)Found97.35 % ofmal Distribution (Conce8\x (d)AXXI (d)81.0 (d)*28.CA (5) AT bris OE (d)88.0 (0)11.0 (d)e2.0 (p)0 (d)A8 IV (9)tsszonow I asW169wo way noZ~ N CM Gnwo muoy no olwo way no onwo uby nonwo wuoy no o 98aro (6) Ta1.ea bns 48 23 (1)ماما ،.281.377. The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation ofThen X~ N(6₂8three. Let X = golf score for a random school team, then fill in the blanks:31

we have to fill blanks. z follows standard normal distribution. here we note that area under…

Answered: 2.5 (9) Week 7 worksheet 6. For a…

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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Heights of a large group of individuals from a certain herbal species is approximately Normally distributed with a mean and standard deviation of 7 and 0.7 inches, respectively. Measurements are rounded to the nearest 0.1 inch. What is the proportion of individuals whose lengths are less than or equal to 7.3 inches? [Refer to appropriate distribution table if needed.] O A. 0.6777 OB. 0.6638 OC. 0.6915 O D. 0.3085
This question pertains to the normal distribution. Part 1 of 3 Assuming that averages on a statistics exam are normally distributed, arrange the following in ascending order. i. an average with a percentile rank of 76% ii. an average at the upper (3rd) quartile iii. an average with a z-score of 0.75 This answer has not been graded yet.
Pumpkins from a local farmer have a mean weight of 16 pounds with a standard deviation of 2.5 pounds. What is the z-score of an 21 pound pumpkin? z=-2 z=+2 z=-1 z=+1
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You are the manager of a restaurant that delivers pizza to customers. You have just changed your delivery process in an effort to reduce the mean time between the order and completion of delivery from the current 25 minutes. From past experience, you can assume that the population standard deviation is 6 minutes. A sample of 36 orders using the new delivery process yields a sample mean of 22.29 minutes. Complete parts (a) to (d) below. Question content area bottom Part 1 a. Using the critical value approach, at the 0.05 level of significance, is there evidence that the mean delivery time has been reduced below the previous population mean value of 25 minutes? Let μ be the population mean. State the null hypothesis, H0, and the alternative hypothesis, H1. A. H0: μ≤25 H1: μ>25 B. H0: μ≥25 H1: μ<25 C. H0: μ≠25 H1: μ=25 D. H0: μ=25 H1: μ≠25 Part 2 The level of significance is 0.05 and the sample size…
Use the following frequency distribution table to find the value of the mean score. Scores Frequency 1-5 3 6-10 6 11-15 7 02.7 09.3 09.8 O 5.3 08.0 Submit Question DII R F F4 % 5 T G J F5 A 6 H F6 & 7 PrtScn 8 FB K Home 9 F9 O End ) 0 F10 ^C P
Im confused on #1.) Letter C. I already found the 68% range and the 95% range by using the empirical rule. Now it is asking " What percentage of this sample fall bellow 90? With the mean of 78 and standard deviation of 6.
Use the standand normal distribution table to determine the percent of data between z= -1.91 and z 1.17 The percent of data between z= -1.91 and z = 1.17 is what?
a. What does 1 standard deviation mean? b. What percentage of scores are falling between the mean and 1 standard deviation?
Heights of boys in a high school are approximately normally distributed with a mean of 175 cm and a standard deviation of 5 cm. What is the first quartile (25%) of heights? (Hint. you need to find the Z-score associated with 25th percentile from the table on page 410) a. 165.8 cm b. 169.6 cm c. 171.7 cm d. 173.5 cm e. 181.0 cm
18
Please complete the following exercises according to Areas under the normal curve table. In an IQ (Intelligence Quotient)distribution (which is a normal distribution)with mean of 100 and a standard deviation of 15, find the percentage of individuals who has and IQ of:a)More than 135(genius)b)Less than 90c)Between 75 and 125

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2.5 (9) Week 7 worksheet 6. For a standard normal distribution (2), answer the following: zor DE.IS equor $8 a. Fill in the blanks. 04 b. P(Z > 1.24) 29tunima.II (b) (@c. P(Z < 2.13) C. 50.05 2=X-M WAB (b) a2.0 (b) PEA.38 (b) d. P(-0.45 c) = 0.39. X=4+2(0) smol f. Find c if P(-c