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Lynna [10]
3 years ago
15

If the random variable X is normally distributed with mean of 50 and standard deviation of 7, find the 9th percentile.

Mathematics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

The 9th percentile is 40.52.

Step-by-step explanation:

We are given that the random variable X is normally distributed with a mean of 50 and a standard deviation of 7.

Let X = <u><em>the random variable</em></u>

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 50

           \sigma = standard deviation = 7

So, X ~ Normal(\mu=50, \sigma^{2} = 7^{2})

<u></u>

<u>Now, the 9th percentile is calculated as;</u>

            P(X < x) = 0.09         {where x is the required value}

            P( \frac{X-\mu}{\sigma} < \frac{x-50}{7} ) = 0.09

            P(Z < \frac{x-50}{7} ) = 0.09

Now, in the z table the critical value of x that represents the below 9% of the area is given as -1.3543, i.e;

                     \frac{x-50}{7}=-1.3543

                     x-50=-1.3543 \times 7

                     x=50 -9.48

                      x = 40.52

Hence, the 9th percentile is 40.52.

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Let x represent the number of female students that are left handed. There are 5 times as many right handed female students as there are left handed female students. This means that the number of female students that are right handed is 5x

Let y represent the number of male students that are left handed. There are 9 times as many right handed male students as there are left handed male students. This means that the number of male students that are right handed is 9x

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