CXKAssume that women's heights are normally distributed with a mean given by μ = 62.6 in, and a standard deviation given by a = 2.4 in.(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.(b) If 42 women are randomly selected, find the probability that they have a mean height less than 63 in.View an example@2F2WS(a) The probability is approximately.(Round to four decimal places as needed.)XHmand#3Ask my instructor80F3EDC$4aF4RF07 dºVTG6MacBook AirBFOYH&7F7ⒸUN*831F81M(9KF9O<IH)OLF10PcommandClear all. V-...IF11{+[=Check ans?option11Show AF12}

Solution: Let X denote the women's height are normally distributed with mean μ=62.6 in and standard…

Answered: Assume that women's heights 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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The heights of 5-year-old children are approximately normally distributed with average height of 100cm and standard deviation of 6cm. Using the emipiracal rule i am trying to figure out What percentage of 5-year-olds are between 90cm and 120cm?
A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.5. Suppose a random sample of 100 male students is selected and the GPA for each student is calculated. Find the interval that contains 95.44 percent of the sample means for male students.
Suppose that the weight of an newborn fawn is Uniformly distributed between 2.2 and 3.9 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. The mean of this distribution is . The standard deviation is . The probability that fawn will weigh exactly 2.5 kg is P(x=2.5)=P(x=2.5)= . The probability that a newborn fawn will be weigh between 2.6 kg and 3.7 kg is P(2.6<x<3.7)=P(2.6<x<3.7)= . The probability that a newborn fawn will be weigh more than 3.24 kg is P(x>3.24)=P(x>3.24)= . P(x>2.8∣x<3.3)=P(x>2.8∣x<3.3)= . Find the 45th percentile. kg
A survey collected data from a random sample of 121 people living in Jade city. The sample average of the distance people travel to reach their workplaces (Y) is 23.28 km and the standard deviation (sy) is 7.48 km. The standard error of the sample average of the distance people travel to reach their workplaces is 68 km. (Round your answer to two decimal places.) Let μy denote the mean of the distance all the people in Jade city travel to reach their workplaces. The p-value of the test Ho: #y #22 km vs. H₁: Hy #22 km is (Round your answer to two decimal places.)
The amount of liquid in a particular group of bottles is normally distributed. The mean is 1.95 L and the standard deviation is 0.15L. If a bottle is chosen at random, find the probability that the amount of water is between 1.88 L and 2.15 L Show all work as in the lessons. Paragraph B I U = I ***
The length of eels (in cm) in a river may be assumed to be normally distributed with a mean of µ = 42 and a standard deviation of o = 6. An angler catches an eel from a river. Let: X = the length (in cm) of an eel a) If the normal length of eels is between 30 cm and 45 cm, what percentage of eels fall within this normal range? Round your answer to 2 decimal places. b) What is the probability that an eel is at least 51 cmin length? Round your answer to 4 decimal places. c) The middle 50% of the lengths (in cm) of the eels are between and Round your answers to 2 decimal places d) Due to the symmetry of the normal distribution we know that DIV 20 - DIV -
A person's body mass index (BMI) is computed by dividing the weight (kg) by the square of height (m). The accompanying table contains the BMI statistics for random samples of males and females. Assume that the two samples are independent simple random samples selected from normally distributedpopulations, and do not assume that the population standard deviations are equal. Let population 1 be females. Female BMI: n=69 x=29.14 s=7.41 Male BMI: n=79 x=28.18 s=5.24 The test statistic is The P-value is Construct a confidence interval appropriate for testing the claim in part (a). The ___% confidence interval estimate is ___<μ1−μ2<___.
Suppose that the mean systolic blood pressure for women over age seventy is 131 mmHg (millimeters of mercury), with a standard deviation of 9 mmHg. Suppose that the blood pressures are normally distributed. Complete the following statements. (a) Approximately? mmHg and 140 mmHg. of women over seventy have blood pressures between 122 (b) Approximately 95% of women over seventy have blood pressures between mmHg and mmHg.
Consider a normal distribution curve where 70-th percentile is at 18 and the 35-th percentile is at 1. Use this information to find the mean, μ, and the standard deviation, o, of the distribution. a) b) o=
Assume that females have pulse rates that are normally distributed with a mean of μ = 76.0 beats per minute and a standard deviation of a = 12.5 beats per minute. Complete parts (a) through (c) below. @ a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 69 beats per minute and 83 beats per minute. The probability is (Round to four decimal places as needed.) b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean between 69 beats per minute and 83 beats per minute. W 6 The probability is (Round to four decimal places as needed.) c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? SA F2 A. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size. OB. Since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample…
Assume for a moment that the distribution of weight of adult is approximately normal with a mean of 179.8 lb and a standard deviation of 93.0 lb find the probably that a randomly selected adult would have a weight at most 272.8 lb A) 0.9413 B) 0.2583 C) 0.8413 D) 0.1583

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