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Answer:
The standard deviation of the scores on a recent exam is 6.
The sample size required is 25.
Step-by-step explanation:
Let <em>X</em> = scores of students on a recent exam.
It is provided that the random variable <em>X</em> is normally distributed.
According to the Empirical rule, 99.7% of the normal distribution is contained in the range, <em>μ </em>± 3<em>σ</em>.
That is, P (<em>μ </em>- 3<em>σ </em>< <em>X</em> < <em>μ </em>+ 3<em>σ</em>) = 0.997.
It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.
This implies that:
P (51 < <em>X</em> < 87) = 0.997
So,
<em>μ </em>- 3<em>σ </em>= 51...(i)
<em>μ </em>+ 3<em>σ </em>= 87...(ii)
Subtract (i) and (ii) to compute the value of <em>σ</em> as follows:
<em> μ </em>- 3<em>σ </em>= 51
(-)<em>μ </em>+ (-)3<em>σ </em>= (-)87
______________
-6<em>σ </em>= -36
<em>σ</em> = 6
Thus, the standard deviation of the scores on a recent exam is 6.
The (1 - <em>α</em>)% confidence interval for population mean is given by:
The margin of error of this interval is:
Given:
MOE = 2
<em>σ</em> = 6
Confidence level = 90%
Compute the <em>z</em>-score for 90% confidence level as follows:
*Use a <em>z</em>-table.
Compute the sample required as follows:
Thus, the sample size required is 25.