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aivan3 [116]
3 years ago
14

A distribution with µ = 55 and σ = 6 is being standardized so that the new mean and standard deviation will be µ = 50 and σ = 10

. when the distribution is standardized, what value will be obtained for a score of x = 52 from the original distribution?
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Solution:

\hbox{Let the distribution standarized is z}\\ \hbox{given:-}\mu =50\hbox{ and} \sigma =6\hbox{will become } \mu=50 \hbox{ and} \sigma =10\\\hbox{distribution standarized with value x=52 }\\\rm{so}\\\hbox{the value of score is:}\\z=\frac{x-\mu }{\sigma}\\\rm{if}\\p(x< 52)=p(z

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Answer:

a_n = 16(\frac{1}{4})^{n - 1}

Step-by-step explanation:

Given:

Fifth term of a geometric sequence = \frac{1}{16}

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Required:

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Solution:

Step 1: find the first term of the sequence

Formula for nth term of a geometric sequence = ar^{n - 1}, where:

a = first term

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Thus, we are given the 5th term to be ¹/16, so n here = 5.

Input all these values into the formula to find a, the first term.

\frac{1}{16} = a*\frac{1}{4}^{5 - 1}

\frac{1}{16} = a*\frac{1}{4}^{4}

\frac{1}{16} = a*\frac{1}{256}

\frac{1}{16} = \frac{a}{256}

Cross multiply

1*256 = a*16

Divide both sides by 16

\frac{256}{16} = \frac{16a}{16}

16 = a

a = 16

Step 2: input the value of a and r to find the nth term formula of the sequence

nth term = ar^{n - 1}

nth term = 16*\frac{1}{4}^{n - 1}

a_n = 16(\frac{1}{4})^{n - 1}

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