Suppose a normal distribution has a mean of 62 and a standard deviation of
4. What is the probability that a data value is between 56 and 64? Round your
answer to the nearest tenth of a percent.

Respuesta :

Answer:

62.5%

Step-by-step explanation:

(56 - 62) / 4 = -1.5 z score = 0.0668

(64 - 62) / 4 = 0.5 z score = 0.6915

0.6915 - 0.0668 = 0.6247

Rounded this becomes 62.5%

The probability for the data value in between 56 and 64 is 62.5%.

What is z-score?

The z-score of a value is the count of the numbers of standard deviations between the value and the mean of set.

How to calculate z-score?

z-score can be calculated by the formula

z = (x - μ)/σ

where,

μ is the mean

σ is the standard deviation

x is the data point

What is probability?

A probability is a number that reflects the chance or likelihood that a particular event will occur.

According to the given question

Mean of a normal distribution, μ = 62

standard deviation of a normal distribution, σ = 4

Now for the data value 56, the z-score will be

z = (56 - 62) / 4 = -1.5 z score = 0.0668

So, the area under normal distribution curve z = -1.5

P(z = -1.5) = 0.0668

Similarly,

For the data value 64, the z-score will be

z-score  = (64 - 62) / 4 = 0.5

The area under normal distribution curve z = 0.5 is given by

P( z = 0.5) = 0.69

Therefore, the probability for the data value in between 56 and 64 is given by

0.6915 - 0.0668 = 0.6247 =  62.5%

Hence, the probability for the data value in between 56 and 64 is 62.5%.

Learn more about the z-score and probability here:

https://brainly.com/question/25638875

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