Assume that head sizes (circumference) of new recruits in the Canadian armed forces can be
approximated by a normal distribution with a mean of 22.8 inches and a standard deviation
of 1.1 inches.
(a)What proportion of recruits have head sizes between 22 and 23 inches?
(b)Five percents of the head sizes exceed _______ inches
To find the proportion of recruits with head sizes between 22 and 23 inches, we can use the concept of the standard normal distribution.
(a) Proportion of recruits with head sizes between 22 and 23 inches:
To solve this, we will first convert the given values to Z-scores using the formula: Z = (x - μ) / σ, where "x" is the value, "μ" is the mean, and "σ" is the standard deviation.
For 22 inches:
Z1 = (22 - 22.8) / 1.1
For 23 inches:
Z2 = (23 - 22.8) / 1.1
Now, we need to find the area under the normal distribution curve between these two Z-scores. This can be done using a standard normal distribution table or a calculator.
Alternatively, we can use a Z-table or a calculator to find the cumulative probabilities associated with each Z-score. Subtracting the smaller cumulative probability from the larger one will give us the proportion of recruits with head sizes between 22 and 23 inches.
(b) To find the head size that exceeds 5% of the recruits, we need to find the Z-score corresponding to the 5th percentile. This can be done by reverse lookup in a Z-table or using a calculator.
Once we have the Z-score, we can convert it back to the actual head size by using the formula: x = Z*σ + μ, where "Z" is the Z-score, "σ" is the standard deviation, and "μ" is the mean.
Using the steps mentioned above, you can find the answers to both parts of the question.
To solve this problem, we will use the properties of the normal distribution.
(a) To find the proportion of recruits with head sizes between 22 and 23 inches, we need to calculate the area under the normal curve between these two values.
We can do this by calculating the Z-scores for the two values and then using a standard normal distribution table or calculator to determine the proportion.
The Z-score for 22 inches is calculated as:
Z1 = (22 - mean) / standard deviation = (22 - 22.8) / 1.1
Similarly, the Z-score for 23 inches is calculated as:
Z2 = (23 - mean) / standard deviation = (23 - 22.8) / 1.1
Using these Z-scores, we can find the proportion of recruits with head sizes between 22 and 23 inches.
(b) To find the head size that exceeds 5% of the recruits, we need to find the Z-score corresponding to the upper 5% tail area of the normal distribution and then convert it back to the original measurement units.
Let's calculate these step-by-step.
(a) Calculation:
Z1 = (22 - 22.8) / 1.1 = -0.7273
Z2 = (23 - 22.8) / 1.1 = 0.1818
Now, we can use these Z-scores to find the proportion of recruits with head sizes between 22 and 23 inches.
P(22 < X < 23) = P(Z1 < Z < Z2)
By referring to a standard normal distribution table or using a calculator, we can find the corresponding probabilities for these Z-scores. Subtracting the lower probability from the higher probability gives us the proportion.
Alternatively, we can use a calculator or statistical software to calculate this directly. Using a normal distribution calculator, we find that the probability is approximately 0.262.
Therefore, approximately 26.2% of recruits have head sizes between 22 and 23 inches.
(b) Calculation:
To find the head size that exceeds 5% of the recruits, we need to find the Z-score representing the upper tail of 5% of the normal distribution. This is denoted as Z(0.05).
Using a standard normal distribution table or calculator, we can find the Z-score associated with the 95th percentile or 0.05 significance level.
Z(0.05) ≈ -1.645
Now, we can convert this Z-score back to the original measurement units to find the head size that exceeds 5% of the recruits.
Value = Z(0.05) * standard deviation + mean
= -1.645 * 1.1 + 22.8
This gives us a value of approximately 20.97 inches.
Therefore, approximately 5% of the recruits have head sizes exceeding 20.97 inches.