A population of values has a normal distribution with µ = 15.7 and s = 1.5

A population of values has a normal distribution with µ = 15.7 and s = 1.5. You intend to draw a random sample of size n = 18. First calculate z, round it to two (2) decimal places, then use the rounded z-score to determine the required probability accurate to four (4) decimal places. Find the probability that a single randomly selected value is less than 16.7. P(x < 16.7) = ?

The correct answer and explanation is :

To determine the probability that a single randomly selected value from a normally distributed population with a mean (µ) of 15.7 and a standard deviation (σ) of 1.5 is less than 16.7, we can follow these steps:

1. Calculate the Z-score:

The Z-score represents how many standard deviations a particular value (X) is away from the mean. It is calculated using the formula:

[ Z = \frac{X – \mu}{\sigma} ]

Given:

  • ( X = 16.7 )
  • ( \mu = 15.7 )
  • ( \sigma = 1.5 )

Substitute these values into the formula:

[ Z = \frac{16.7 – 15.7}{1.5} = \frac{1}{1.5} \approx 0.6667 ]

Rounded to two decimal places, the Z-score is:

[ Z \approx 0.67 ]

2. Determine the probability corresponding to the Z-score:

The Z-score of 0.67 indicates that the value 16.7 is 0.67 standard deviations above the mean. To find the probability that a randomly selected value is less than 16.7, we need to determine the cumulative probability associated with a Z-score of 0.67.

Using a standard normal distribution table or a Z-score calculator, we find that the cumulative probability for a Z-score of 0.67 is approximately 0.7486.

3. Interpret the result:

A cumulative probability of 0.7486 means that there is a 74.86% chance that a randomly selected value from this population will be less than 16.7.

Explanation:

The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, symmetric around the mean. The Z-score standardizes a value, allowing us to determine its position relative to the mean in terms of standard deviations. By converting the raw score to a Z-score and consulting the standard normal distribution, we can find the probability of observing a value less than or equal to the given score.

The shaded area under the curve to the left of Z = 0.67 represents the cumulative probability of approximately 74.86%.

Conclusion:

By calculating the Z-score for the value 16.7 and determining the corresponding cumulative probability, we find that there is a 74.86% probability that a single randomly selected value from this normal distribution will be less than 16.7.

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