If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen ts have heights (a) greater than 72 inches, (b) less than or equal to 64 inches, (c) between 65 and 71 inches inclusive, (d) equal to 68 inches? assume the measurements to be recorded to the nearest inch.
1 answer:
Given: μ = 68 in, population mean σ = 3 in, population standard deviation Calculate z-scores for the following random variable and determine their probabilities from standard tables. x = 72 in: z = (x-μ)/σ = (72-68)/3 = 1.333 P(x) = 0.9088 x = 64 in: z = (64 -38)/3 = -1.333 P(x) = 0.0912 x = 65 in z = (65 - 68)/3 = -1 P(x) = 0.1587 x = 71: z = (71-68)/3 = 1 P(x) = 0.8413 Part (a) For x > 72 in, obtain 300 - 300*0.9088 = 27.36 Answer: 27 Part (b) For x ≤ 64 in, obtain 300*0.0912 = 27.36 Answer: 27 Part (c) For 65 ≤ x ≤ 71, obtain 300*(0.8413 - 0.1587) = 204.78 Answer: 204 Part (d) For x = 68 in, obtain z = 0 P(x) = 0.5 The number of students is 300*0.5 = 150 Answer: 150
You might be interested in
14 = 0.5p + 12 <== ur equation
Answer:
2x+8-2y
Step-by-step explanation:
4x-2x+5+3-2y
2x+8-2y
54 divided by 6 would be 9 meaning he takes 9 quizzes a week. If directly proportional by the end of 7 weeks he would have taken 63 quizzes
B b b b b b b b b b b b b b b b b b b b b b b b b b b b b b.
Answer:
There is any answer as you choose.
Step-by-step explanation:
There is no equal to so you can choose any number.