Assume that SAT scores are normally distributed with a mean mu = 1518 and standa

Question

Assume that SAT scores are normally distributed with a mean mu = 1518 and standard deviation sigma = 325. If 1 SAT score is randomly selected, find the probability that it is less than 1500. Calculate z score and use Table A-2 to determine the probability p, z = (x - mu)/sigma use two decimal places for; score Answer: ________ Assume that SAT scores are normally distributed with a mean mu = 1518 and standard deviation sigma = 325. If 100 SAT score is randomly selected, find the probability that it is less than 1500. Calculate the new standard deviation sigma_x, = sigma/squarerootof n, the z score and use Table A-2 to determine the probability p. (z = (x - mu)/sigma use two decimal places for z score) Answer ______ How high should doorways be if they are designed for 95% of the men? Mean height for men is mu = 69 inches, and the standard deviation is sigma = 2.8 inches. From Table A-2 we find that the z score for .9500 is 1.645. Find the optimum height for doorways (x) using the formula x = mu + (z + sigma). Answer: __________ Membership in Mensa requires an IQ score above 131. IQ scores are normally distributed with a mean mu = 100 and standard deviation sigma = 15. If 1 person is randomly selected from the general population, find the probability of getting someone with an IQ greater than 135. Use Table A-2 for area under the curve, and z = (x - mu)/sigma. (two decimal places for z score) Answer: ______ The distribution of sample x will, as the sample size increases, approach a normal distribution. True False

Explanation / Answer

7)
mean = 1518 , s = 325
P(X < 1500)
z = ( x - mean) / s
= ( 1500 - 1518) / 325
= -0.055
we need to find P( Z < -0.055)
P(X < 1500) = P( Z < -0.055) = 0.4779

8)

mean = 1518 , s = 325 , n = 100
P(X < 1500)
z = ( x - mean) / (s/sqrt(n))
= ( 1500 - 1518) / (325/sqrt(100))
= -0.5538

we need to find P( Z < -0.5538)
P(X < 1500) = P( Z < -0.5538) = 0.2897

9)

z score = 1.645 , mean = 69 , s = 2.8
x = mean + z * s
= 69 + 1.645*2.8
= 73.606

10)
mean = 100 , s = 15
P(X > 135)
z = ( x - mean) / s
= ( 135 - 100) / 15
= 2.333

we need to find P( Z >2.333)
P(X > 135) = P( Z > 2.333) = 0.0098

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