A normal distribution has a mean of 32 and a standard deviation of 4. Find the probability that a randomly selected xx -value from the distribution is at most 35. Round to four decimal places.

cistG

cistG

Answered question

2021-02-03

The average value of a normal distribution is 32, and the standard deviation is 4. Determine the likelihood that a randomly chosen xx-value from the distribution is no greater than 35. Four decimal places should be used.

Answer & Explanation

opsadnojD

opsadnojD

Skilled2021-02-04Added 95 answers

To find the probability of a selected x-value being at most 35, you need to find the corresponding z-score and then use a z-score table. The area below the curve for the z-score is then the probability that the value will be at most 35.
To find the corresponding z-score, use the formula z=xμσ where μμ is the mean and σ is the standard deviation.

Since you want the value to be at most x=35, the mean is μ=32, and the standard deviation is σ=4, then z=35−324=34=0.75.

Using a z-score table gives an area of 0.7734 for a z-score of z=0.75. You can also calculate the area using the normalcdf feature on a graphing calculator.

Therefore, the probability of a selected x-value being at most 35 is 0.7734.

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