The heights of boys at a particular age follow a normal distribution with mean 150.3 and standard deviation 5cm. Find the probability that the height of a boy picked at random from the age group is i.more than 10 cm from the mean height.
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To find the probability that the height of a boy picked at random from the age group is more than 10 cm from the mean height, we can use the standard normal distribution.
Step 1: Convert the problem into the standard normal distribution by converting the values to z-scores.
To find the z-score, we can use the formula:
z = (x - μ) / σ
where:
x is the observed value,
μ is the mean,
and σ is the standard deviation.
In this case, we have:
x = 150.3 + 10 (since we want the height to be more than 10 cm from the mean)
μ = 150.3 (the given mean)
σ = 5 (the given standard deviation)
Substituting the values into the formula, we get:
z = (150.3 + 10 - 150.3) / 5
z = 10 / 5
z = 2
Step 2: Find the probability using the standard normal distribution table or a calculator.
To find the probability corresponding to the z-score of 2, we can use a standard normal distribution table or a calculator.
Using a standard normal distribution table, we can find that the probability of the z-score being greater than 2 is approximately 0.0228.
So, the probability that the height of a boy picked at random from the age group is more than 10 cm from the mean height is approximately 0.0228 or 2.28%.
To find the probability that the height of a boy picked at random from the age group is more than 10 cm from the mean height, we need to calculate the area under the normal distribution curve beyond 10 cm from the mean.
First, we need to find the z-score corresponding to 10 cm from the mean. The z-score, also known as the standard score, measures the number of standard deviations an observation or data point is from the mean. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the value of interest (in this case, 10 cm)
- μ is the mean of the distribution (150.3 cm)
- σ is the standard deviation of the distribution (5 cm)
Plugging in the values, we get:
z = (10 - 150.3) / 5
z = -140.3 / 5
z = -28.06
Next, we need to find the area under the normal distribution curve beyond the z-score of -28.06. Since the normal distribution is symmetric, the area beyond a negative z-score is the same as the area beyond the corresponding positive z-score.
Using a standard normal distribution table or a calculator, we can find that the area to the right of a z-score of -28.06 is approximately 0 (to several decimal places). Therefore, the probability that the height of a boy picked at random from the age group is more than 10 cm from the mean height is almost 0.