
How do I find the range of lengths in this question? The length of a Dachshund is normally distributed with a mean of 15 inches and a standard deviation of 3.5 inches. What range of lengths would occur 75% of the time?
Answer
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Hint: In a normal distribution around 75% of the values lie within 1.15 S.D units from the population mean. This can be used to find the range of lengths that would occur 75% of the time.
Complete step by step solution:
We are given a normal population and are asked to find the range of lengths that would occur 75% of the time. This means that a randomly chosen length from the population lies in the required range 75% times. That means that the probability of choosing a length in the range is 75%. So, we have to find an interval around the mean, where the area under the normal curve is 75%. We will use the standard normal tables to find such an interval.
Here, the population is normally distributed with mean \[\mu = 15\] and standard deviation $ \sigma = 3.5 $ .
Now, we know from the standard normal tables the $ Z $ value that corresponds to 75% of the area under the curve is 1.15. This means that around 75% of the values lie within 1.15 $ \sigma $ units from the mean $ \mu $ . This result comes from the transformation applied to get $ Z $ , which is $ Z = \dfrac{{X - {\mu _X}}}{{{\sigma _X}}} $ , where $ X $ denotes the values.
$ \Rightarrow X = {\mu _X} \pm Z{\sigma _X} $
Now, since we know the values of $ {\mu _X}{\text{ and }}{\sigma _X} $ , we can use the above result to find the range of values.
$ \Rightarrow X = 15 \pm 1.15 \times 3.5 $
$ \Rightarrow X = 15 \pm 4.025 $
So, around 75% values would lie in $ \left( {10.975,19.025} \right) $ .
Hence the range of lengths that would occur 75% of the time is between $ 10.975 $ inches and $ 19.025 $ inches.
Note: It is important to know how to interpret the values given in the standard normal table, and also to find the area under the curve for a given $ Z $ value and vice-versa. Interpretation of the results after solving the problem is also important in statistics. Use Chebyshev’s inequality, if the population is not normal.
Complete step by step solution:
We are given a normal population and are asked to find the range of lengths that would occur 75% of the time. This means that a randomly chosen length from the population lies in the required range 75% times. That means that the probability of choosing a length in the range is 75%. So, we have to find an interval around the mean, where the area under the normal curve is 75%. We will use the standard normal tables to find such an interval.
Here, the population is normally distributed with mean \[\mu = 15\] and standard deviation $ \sigma = 3.5 $ .
Now, we know from the standard normal tables the $ Z $ value that corresponds to 75% of the area under the curve is 1.15. This means that around 75% of the values lie within 1.15 $ \sigma $ units from the mean $ \mu $ . This result comes from the transformation applied to get $ Z $ , which is $ Z = \dfrac{{X - {\mu _X}}}{{{\sigma _X}}} $ , where $ X $ denotes the values.
$ \Rightarrow X = {\mu _X} \pm Z{\sigma _X} $
Now, since we know the values of $ {\mu _X}{\text{ and }}{\sigma _X} $ , we can use the above result to find the range of values.
$ \Rightarrow X = 15 \pm 1.15 \times 3.5 $
$ \Rightarrow X = 15 \pm 4.025 $
So, around 75% values would lie in $ \left( {10.975,19.025} \right) $ .
Hence the range of lengths that would occur 75% of the time is between $ 10.975 $ inches and $ 19.025 $ inches.
Note: It is important to know how to interpret the values given in the standard normal table, and also to find the area under the curve for a given $ Z $ value and vice-versa. Interpretation of the results after solving the problem is also important in statistics. Use Chebyshev’s inequality, if the population is not normal.
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