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the results of a certain medical test are normally distributed with a m…

Question

the results of a certain medical test are normally distributed with a mean of 129 and a standard deviation of 15. convert the given results into z - scores, and then use the accompanying table of z - scores and percentiles to find the percentage of people with readings between 153 and 165. click the icon to view the table of z - scores and percentiles. the percentage of people with readings between 153 and 165 is \\( \square \\% \\). (round to two decimal places as needed.)

Explanation:

Step1: Calculate z - score for \(x = 153\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 129\) (mean) and \(\sigma=15\) (standard deviation).
For \(x = 153\), \(z_1=\frac{153 - 129}{15}=\frac{24}{15}=1.6\)

Step2: Calculate z - score for \(x = 165\)

Using the same formula \(z=\frac{x-\mu}{\sigma}\)
For \(x = 165\), \(z_2=\frac{165 - 129}{15}=\frac{36}{15}=2.4\)

Step3: Find the corresponding percentiles

Looking up in the z - score table:
The percentile for \(z = 1.6\) is \(0.9452\) (this means that approximately \(94.52\%\) of the data is less than \(z = 1.6\))
The percentile for \(z = 2.4\) is \(0.9918\) (this means that approximately \(99.18\%\) of the data is less than \(z = 2.4\))

Step4: Calculate the percentage between the two z - scores

The percentage of data between \(z_1 = 1.6\) and \(z_2=2.4\) is \(P(1.6<Z<2.4)=P(Z < 2.4)-P(Z < 1.6)\)
\(P(1.6<Z<2.4)=0.9918 - 0.9452=0.0466\)

Answer:

\(4.66\%\)