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quester [9]
4 years ago
10

Refer to the following information from a frequency distribution for "heights of college women" recorded to the nearest inch:The

first two class midpoints are 62.5" and 65.5".What are the class limits for the lowest class? A. 61 and up to 64B. 62 and up to 64C. 62 and 65D. 62 and 63
Mathematics
1 answer:
valkas [14]4 years ago
4 0

Answer: Therefore, Option 'A' is correct.

Step-by-step explanation:

Since we have given that

Two class midpoints are 62.5 and 65.5

So, the class interval would be

65.5-62.5=3

So, the limits of lower class are x_1\ and\ x_2

The limits of upper class are x_2\ and\ x_3

So, it becomes,

\dfrac{x_1+x_2}{2}=62.5\\\\x_1+x_2=62.5\times 2=125

and

\dfrac{x_2+x_3}{2}=65.5\\\\x_2+x_3=65.5\times 2=131

Since we have given that interval is 3.

So, x_2-x_1=3\ and\ x_3-x_2=3

So, by solving all the questions, we get that

x_2+x_1=125\\\\x_2-x_1=3\\\\-------------------------\\\\2x_2=128\\\\x_2=64\\\\x_2-x_1=3\\\\64-x_1=3\\\\x_1=61\\\\and\\\\x_3-x_2=3\\\\x_3-64=3\\\\x_3=67

Hence, the class limits would be

(61,64) and (64,67).

And the class limits of lowest class are 61 and 64.

Therefore, Option 'A' is correct.

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1st Step: Started off by regrouping the terms
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3rd Step: Now we can simplify the 1/9 to just x^4x^2/9

4th Step: Now we can use the product rule which is simple. So We add the exponents and simplify it to just one exponent. So x4+2=6 that simplifies to just x^6.

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Step-by-step explanation:

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Answer:

7.3% of the bearings produced will not be acceptable

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 0.499, \sigma = 0.002

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.504 - 0.494}{0.002}

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pvalue of Z when X = -1.5

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Z = -1.5

Z = -1.5 has a pvalue of 0.0668

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7.3% of the bearings produced will not be acceptable

4 0
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