Q.N.9) The board of examiners the found that the mean score on the test was 426

Question

Q.N.9) The board of examiners the found that the mean score on the test was 426 and the standard deviation was 72. 1 set the passing scor that the scores are normally distributed.? that administers the real estate broker's examination in a certain s f the board wants to Assume re so that only the best 80% of all applicants pass, what is the passing score? Q N. 10) A random variable X has a normal distribution with mean 25 and standard deviation 5, Find the probability that a) X lies between 24 and 26. b) X is greater than 40. c) X is exactly 25. Extra Credit: Body index and 2 A study by the of women in this age range is overweight y mass index (BMI) is equal to "weight in kilograms" divided by "height is meters ny by the National Center for health Statistics suggested that women between the ages 20 ed States have a mean BMI of 26.8 with a standard deviation of 7.4. Let us assume that distributed. A BMI of 30 or greater is classified as being overweight. What proportion

Explanation / Answer

Question 9

We are given

Mean = 426

SD = 72

We have to find passing score X for upper 80% scores.

X follows normal distribution.

X = Mean + Z*SD

Z for upper 80% or lower 20% is given as below:

Z = -0.841621234

X = 426 + (-0.841621234)*72

X = 426 – 0.841621234*72 = 365.4032712

Passing score = 365.40

Question 10

X follows normal distribution.

Mean = 25

SD = 5

Part a

P(24<X<26) = P(X<26) – P(X<24)

Z = (X – Mean) / SD

For X<26

Z = (26 – 25) / 5 =1/5 = 0.20

P(Z<0.20) = P(X<26) = 0.579259709

For X<24

Z = (24 – 25)/5 = -1/5 = -0.20

P(Z<-0.20) = P(X<24) = 0.420740291

P(24<X<26) = P(X<26) – P(X<24)

P(24<X<26) = 0.579259709 - 0.420740291

P(24<X<26) = 0.158519419

Required probability = 0.158519419

Part b

P(X>40) = 1 – P(X<40)

Z = (40 – 25) / 5 = 15/5 = 3

P(Z<3) = P(X<40) = 0.99865

P(X>40) = 1 – P(X<40) = 1 – 0.99865 =0.00135

Required probability = 0.00135

Part c

P(X=25) = 0.00

(Exact probabilities for continuous distribution are always zero.)

Extra Credit:

Mean = 26.8

SD = 7.4

We have to find P(X>30)

P(X>30) = 1 – P(X<30)

Z = (30 – 26.8) / 7.4 = 0.432432432

P(Z< 0.432432432) = P(X<30) = 0.667286425

P(X>30) = 1 – P(X<30)

P(X>30) = 1 – 0.667286425

P(X>30) = 0.332713575

Required probability = 0.332713575

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