the heights of female students at a particular college are normally distributed with the mean 169 cm and standard deviation 9 cm.
1) Given that 80% of these female students have a height less than h cm ,find the value of h.
2) Given that 60% of these female students have a height greater than s cm,find the value of s.
1) H~N(169,9^2)
(H<h)=0.80
(H< (h-169)/9)=0.80
(H<a)=0.80
(fi)a =0.80 (look for 0.80 in the normal distribution table)
the closest # to 0.80 is 0.7995 at 0.84
0.0005 is left to make 0.80 so look for that in the add part of the table 5 is at 2
so the a=0.842
we suggested a as (h-169)/9 =0.842
solve, h=176.6
2) (H>s)=0.6
Standardize the s , (H>(s-169)/9)=0.6
(H>b)=0.6
s is negative because ; if its greater than the random variable which is (s) and the probability is greater than 0.5 then the random variable is negative
rule states that ; (fi)b=0.6
from the table (0.6) is 0.5987 at 0.25
0.0013 left for it to equal 0.6
from the table take the minimum value which is 12 at 3
so b is 0.253
its ngative so b is -0.253
(s-169)/9=-0.253
s=166.7
0.7
1) Well, 80% of the students have a height less than h cm. So, we need to find the z-score corresponding to the 80th percentile.
Using the z-table, we find that the z-score corresponding to the 80th percentile is approximately 0.845.
Now, we can use the formula for z-score:
z = (x - μ) / σ
where x is the value we're looking for, μ is the mean, and σ is the standard deviation.
In this case, we have:
0.845 = (h - 169) / 9
Solving for h, we get:
h - 169 = 0.845 * 9
h - 169 = 7.605
h = 7.605 + 169
h ≈ 176.6 cm
2) Similarly, we need to find the z-score corresponding to the 60th percentile, which is approximately 0.253.
Using the z-score formula, we have:
0.253 = (s - 169) / 9
Solving for s, we get:
s - 169 = 0.253 * 9
s - 169 = 2.277
s = 2.277 + 169
s ≈ 171.3 cm
So, the value of h is approximately 176.6 cm, and the value of s is approximately 171.3 cm.
To solve these problems, we will use the Z-score formula, which allows us to convert normal distribution values into standard units, or Z-scores.
The formula for the Z-score is:
Z = (X - μ) / σ
where:
Z = Z-score
X = Value from the distribution
μ = Mean of the distribution
σ = Standard deviation of the distribution
1) To find the value of h, we need to find the Z-score corresponding to the 80th percentile. In a normal distribution, the area to the left of the Z-score represents the percentage of values that are less than that Z-score.
We can use a Z-score table or a calculator to find the Z-score that corresponds to the 80th percentile. The Z-score is approximately 0.8416.
Now we use the Z-score formula to solve for h:
Z = (h - μ) / σ
0.8416 = (h - 169) / 9
Cross-multiply and solve for h:
0.8416 * 9 = h - 169
h - 169 = 7.5744
h = 7.5744 + 169
h ≈ 176.5744
The value of h is approximately 176.5744 cm.
2) To find the value of s, we need to find the Z-score corresponding to the 60th percentile. We can use a Z-score table or a calculator to find that the Z-score is approximately -0.2533.
Using the Z-score formula again:
Z = (s - μ) / σ
-0.2533 = (s - 169) / 9
Cross-multiply and solve for s:
-0.2533 * 9 = s - 169
-2.2797 = s - 169
s = -2.2797 + 169
s ≈ 166.7203
The value of s is approximately 166.7203 cm.
So, the value of h is approximately 176.5744 cm, and the value of s is approximately 166.7203 cm.
To find the value of h (height) in the first question, we need to determine the corresponding z-score (the number of standard deviations away from the mean) using the standard normal distribution table.
1) Using the standard normal distribution table:
- The area to the left of the z-score represents the proportion of observations below that value.
- We know that 80% of the observations are below h, so we need to find the z-score where the cumulative area to the left is 0.80.
Now let's calculate the z-score for h:
Step 1: Convert the given proportion (80%) to a decimal: 0.80.
Step 2: Subtract this decimal from 1 to determine the area to the right: 1 - 0.80 = 0.20.
Step 3: Find the z-score associated with this right-tailed area of 0.20 (using the standard normal distribution table or calculator).
The z-score is approximately -0.84 (from the table).
Next, we can use the formula for standardizing a value (x) in a normal distribution to find the value of h:
z = (x - mean) / standard deviation
Rearranging the formula, we have:
x = (z * standard deviation) + mean
Plugging in the values from the question:
x = (-0.84 * 9) + 169
x = -7.56 + 169
x = 161.44
Therefore, the value of h is approximately 161.44 cm.
Moving on to the second question:
2) Given that 60% of the observations are greater than s (height in cm), we need to determine the z-score associated with this right-tailed area.
Step 1: Convert the given proportion (60%) to a decimal: 0.60.
Step 2: Subtract this decimal from 1 to determine the area to the left: 1 - 0.60 = 0.40.
Step 3: Find the z-score associated with this left-tailed area of 0.40 (using the standard normal distribution table or calculator).
The z-score is approximately 0.25 (from the table).
Using the formula for standardizing a value (x) in a normal distribution, we can find the value of s:
x = (z * standard deviation) + mean
Plugging in the values from the question:
x = (0.25 * 9) + 169
x = 2.25 + 169
x = 171.25
Therefore, the value of s is approximately 171.25 cm.