if it was XN(6) Between that - the Mean deviation in this distribution is 2 6 T the First Spring is Q-μ- 0.676 And the third Spring is or μ+ 0.676
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- A normal distribution has a mean of µ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 90 and X = 110?If X has normal distribution with mean 80 and standard deviation 15. Find the P(X > 90) a. 0.7734 b. -0.75 Oc. 0.75 O d. 0.2514IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = 1Q of an individual. (a) Find the z-score for an IQ of 97, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 97. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x-axis is the z- Score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value Click and drag the arrows to adjust the values. -1 3 4 -1.5 (d) MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. (e) Sketch the graph, and write the probability statement. BIUX, x' C 次四 Edit. Insert - Formats - Σ ΣΗ
- A population has mean u=30 and standard deviation o=7. (A) Find the z-score for a population value of 9. (B) Find the z-score for a population value of 51. (C) What number has a z-score of -2.3The variable W represents the weight of potatoes buns sold at the bakery. This variable is normally distributed with mean 90 g and standard deviation 6 g. (a) (i) Explain the meaning of S6 . (ii) Find the mean and standard deviation of S6 . (ii) Hence, Calculate P ( S6 2 560 g).From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 38 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? (i) Alpha (Enter an exact number as an integer, fraction, or decimal.)? =
- A normal population has mean =μ9 and standard deviation =σ5 . Find the proportion of the population that is greater than 4 . Round the answers to at least four decimal places. The proportion of the population that is greater than 4 is .The mean is µ = 15.2 and the standard deviation is Η = 0.9.Find the probability that X is between 14.3 and 16.1.X is a normally distributed variable with mean μ = 30 and standard deviation σ = 6. Find the following: (Round to three decimal places) P(x < 40) P(x > 21) P(30 < x < 35)
- A population of values has a normal distribution with μ=219.5μ=219.5 and σ=62.5σ=62.5. A random sample of size n=187n=187 is drawn. What is the mean of the distribution of sample means?μ¯x=μx¯= What is the standard deviation of the distribution of sample means? Round your answer to two decimal places.σ¯x=σx¯=Suppose you simulate the price path of stock HHF using a geometric Brownian motion model with µ = 0.1, o = 0.2, and time step At = 1/52 (weekly). Let St be the price of the stock at time t. If So = 100, and the first simulated (randomly selected) standard normal variable is e = -0.591, what is the simulated return over the next week? (a) -0.061 (b) -0.015 (c) 0.061 (d) -0.093Find P(11<X<15) If X is normally distributed with a mean of 14 and a standard deviation of 2?
