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?

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  1. 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%

  2. 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.

  3. [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.

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