Question

The 16th percentile in a Normal distribution is 25 and 2.5% of the values in the distribution are greater than 40.
Which of the following are the best estimates for the mean and standard deviation of the distribution?
(a) mean = 32.5; standard deviation = 2.5
(b) mean = 32.5; standard deviation 5
(c) mean 32.5; standard deviation = 10
(d) mean = 30; standard deviation - 2.5
(e) mean = 30; standard deviations

Answer 1

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Complete Question

The 16th percentile in a Normal distribution is 25 and 2.5% of the values in the distribution are greater than 40.

Which of the following are the best estimates for the mean and standard deviation of the distribution?

(a) mean = 32.5; standard deviation = 2.5

(b) mean = 32.5; standard deviation = 5

(c) mean 32.5; standard deviation = 10

(d) mean = 30; standard deviation = 2.5

(e) mean = 30; standard deviations = 5

Answer:

(e) mean = 30; standard deviations = 5

Step-by-step explanation:

From the question, we are told that the 16th percentile in a Normal distribution is 25

Hence the standard deviation is calculated as √of the 16th percentile in a normal distribution

= √25

= 5

Hence, the standard deviation = 5

In other to obtain the mean, we used the formula of z score.

z score formula =(x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

From the question

x = 40, σ = 5

We are told that 2.5 % of the values is greater than 40

This means The percentile for this values = 100 - 2.5 %

= 97.5%

The z-score for a 97.5th percentile = 1.96

Hence,

z = (x-μ)/σ

1.96 = (40 - μ) / 5

Cross Multiply

= 1.96 × 5 = 40 - μ

= 9.80 = 40 - μ

μ = 40 - 9.80

μ = 30.2

≈ 30

Hence, the Mean = 30

Answer 2

A normal distribution is symmetric about the mean.

• The mean of the distribution is 30
• The standard deviation is 5

Given

$P_{16} = 25$ ---- 16th percentile

First, we calculate the standard deviation as follows:

$\sigma = \sqrt{P_{16}$

So, we have:

$\sigma = \sqrt{25$

$\sigma = 5$

Next, we have:

$2.5\%$ of the value are greater than 40

This means that:

$x =40$ and 97.5% are less.

So: the z-score of 97.5% is:

$z = 1.960$

Using z formula:

$z = \frac{x - \mu}{\sigma}$

Substitute values for z, x and $\sigma$

So, we have:

$1.960 = \frac{40 - \mu}{5}$

Multiply through by 5

$5 \times 1.960 = 40 - \mu$

$9.8 = 40 - \mu$

Collect like terms

$\mu = 40 - 9.8$

$\mu = 30.2$

Approximate

$\mu = 30$

Hence, the mean is 30 and the standard deviation is 5

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