Question

ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal. Find the standard deviation.

Answer 1

The standard deviation of the ACT scores is found to be approximately 5.37, using the normal distribution properties and z-score for the 91st percentile.

Step-by-step explanation:

To find the standard deviation of ACT scores when given that the mean is 20.8 and 9 percent of the scores are above 28, we need to use the concept of the normal distribution and z-scores.

Since 9 percent of scores are above 28, this corresponds to a z-score that reflects the 91st percentile of a standard normal distribution. We can use z-score tables or a statistical calculator to find that the z-score associated with the 91st percentile is approximately 1.34. Once we have the z-score, we can use the formula z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

Plugging in the values, we get 1.34 = (28 - 20.8) / σ. Solving for σ, we find that the standard deviation is approximately 5.37.

Therefore, the standard deviation of the ACT scores is roughly 5.37.

Answer 2

Answer: 5.37

Step-by-step explanation:

Let x = ACT scores.

Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.

i.e. P(X>28)=0.09 (i)

Now,

`P(X>28)=P((X-\mu)/(\sigma)>(28-20.8)/(\sigma))\\\\= P(z>(7.2)/(\sigma))\ \ \ \ [z=(X-\mu)/(\sigma)]` (ii)

One -tailed z value for p-value of 0.09 =1.3408 [By z-table]

From (i) and (ii)

`(7.2)/(\sigma)=1.3408\\\\\Rightarrow\ \sigma=(7.2)/(1.3408)\\\\\Rightarrow\ \sigma=5.37`

Hence, the standard deviation = 5.37