Anormal distribution has a mean of 103 and a standard deviation of 2.7 what is the z-score of 105?

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Anormal distribution has a mean of 103 and a standard deviation of 2.7 what is the z-score of 105?

Probability = 0.15%

Step-by-step explanation:

The provided information is:

Let X be the scores of the IQ scores that is normally distributed with mean [tex]\mu = 103[/tex] and standard deviation

.

That is [tex]X \sim N(103,15)[/tex]

[tex]103 - 58 = 45 = 3 \times 15[/tex]

So, 58 is 3 standard deviation left to the mean.

As 99.7% of probability lies within the three standard deviation.

Thus, above three standard deviation is:

[tex]\frac{1-0.997}{2}=0.0015[/tex]

Thus, the required probability is 0.15%

z-score = 0.741

Step-by-step explanation:

The provided information are:

Population mean [tex](\mu)[/tex] = 103

Population standard deviation [tex](\sigma)[/tex] = 2.7

Score (X) = 105

The formula to calculate the z-score is:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

The z- score can be calculated as:

[tex]z=\frac{X-\mu}{\sigma}\\z=\frac{105-103}{2.7} \\z= 0.741[/tex]

Hence, the z-score is 0.741.

[tex]P(61[/tex]

And we can use the z score formula in order to find the deviationn above/below for the limits given given by:

[tex]z= \frac{X -\mu}{\sigma}[/tex]

And replacing we got:

[tex]z=\frac{61-103}{14}= -3[/tex]

[tex]z=\frac{145-103}{14}= 3[/tex]

So then we want the % of values within 3 deviation from the mean and from the empirical rule we know that between these we have 99.7% of the data.

Step-by-step explanation:

We know that the IQ scores have the following parameters:

[tex]\mu = 103, \sigma = 14[/tex]

And we want to find the following probability:

[tex]P(61[/tex]

And we can use the z score formula in order to find the deviationn above/below for the limits given given by:

[tex]z= \frac{X -\mu}{\sigma}[/tex]

And replacing we got:

[tex]z=\frac{61-103}{14}= -3[/tex]

[tex]z=\frac{145-103}{14}= 3[/tex]

So then we want the % of values within 3 deviation from the mean and from the empirical rule we know that between these we have 99.7% of the data.

x = 105, xav = 103, sd = 2.7

Dang i thought i had it but i forgot lol