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

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

9. Find the area of a circle having a circumference of 382. Round to the nearest tenth. Use 3.14 for 1. a. 1133.5 units b. 1078.6

1. manuellopez1981 says:

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 $\mu = 103$   and standard deviation

.

That is  $X \sim N(103,15)$

$103 - 58 = 45 = 3 \times 15$

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:

$\frac{1-0.997}{2}=0.0015$

Thus, the required probability is 0.15%

2. nenaa29 says:

z-score = 0.741

Step-by-step explanation:

The provided information are:

Population mean $(\mu)$ = 103

Population standard deviation $(\sigma)$ = 2.7

Score (X) = 105

The formula to calculate the z-score is:

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

The z- score can be calculated as:

$z=\frac{X-\mu}{\sigma}\\z=\frac{105-103}{2.7} \\z= 0.741$

Hence, the z-score is 0.741.

3. simrankaurdhatt says:

$P(61$

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

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

And replacing we got:

$z=\frac{61-103}{14}= -3$

$z=\frac{145-103}{14}= 3$

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:

$\mu = 103, \sigma = 14$

And we want to find the following probability:

$P(61$

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

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

And replacing we got:

$z=\frac{61-103}{14}= -3$

$z=\frac{145-103}{14}= 3$

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.

4. emilyturchon says:

x = 105, xav = 103, sd = 2.7

5. swaggg2084 says:

Dang i thought i had it but i forgot lol