Home Mathematics 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?Mathematics mobeteOctober 22, 20215 CommentsAnormal 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%Reply
z-score = 0.741Step-by-step explanation:The provided information are:Population mean [tex](\mu)[/tex] = 103Population standard deviation [tex](\sigma)[/tex] = 2.7Score (X) = 105The 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.Reply
[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.Reply
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