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Alexxandr [17]
2 years ago
4

The mean score for freshmen on an aptitude test at a certain college is 540, with a standard deviation of 50. Assume the means t

o be measured to any degree of accuracy. What is the probability that two groups selected at random, consisting of 32 and 50 students, respectively, will differ in their mean scores by (a) more than 20 points
Mathematics
1 answer:
torisob [31]2 years ago
6 0

Answer:

P ( (X-Y) > |20| ) = 0.0772

Step-by-step explanation:

Solution:-

- Denote a random variable (X) with mean ux = 540 and standard deviation sx = 50 describes the sample of first group of nx = 32 students from the observed population.

- Denote a random variable (Y) with mean uy = 540 and standard deviation sy = 50 describes the sample of first group of ny = 50 students from the observed population.

- The sampling distribution is approximated normal ( X - Y ), where:

                     u(x-y) = ux - uy = 0

                     s(x-y) = √( sx^2/nx + sy^2/ny) = √( 50^2/32 + 50^2/50) = 11.319

- The distribution of difference in samples is given as:

                    (X-Y) ~ ( 0 , 11.319^2)

a)

- For the difference to be greater than 20.

- We will standardize our distribution (X-Y) to Z-statistics:

                    P ( (X-Y) > |20| ) = P ( (X-Y) < -20 ) + P ( (X-Y) > 20 )

                                              = P ( Z < (-20 - 0) / 11.319) + P (Z > (20 - 0) / 11.319)

                                              = P ( Z < -1.767) +1 - P (Z < 1.767)

                                              = 0.0386 + 1 - 0.9614

                                              = 0.0772

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Answer:

Compatible width of rectangular banquet hall could be 90 feet approximately.

Step-by-step explanation:

Given

Area of a rectangular banquet hall = 7400 square feet

Length of rectangular banquet hall = 82 feet

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Now Area of rectangle is equal to length times width.

Framing in equation form we get;

Area\ of \ Rectangle= length \times width

Substituting the given values we get;

82 \times width = 7400\\\\width = \frac{7400}{82} = 90.2439

Now by definition of Compatible numbers which state that:

"Compatible numbers are the numbers that are easy to add, subtract, multiply, or divide mentally.  Compatible numbers are close in value to the actual numbers that make estimating the answer and computing problems easier."

Now By Using width as 90 feet and length as 82 feet we get area as 7380 sq ft. which is closet to actual area which is 7400 sq ft.

Hence we can say compatible width could be 90 feet approximately.

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2 years ago
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