QUESTION IMAGE
Question
- the wechsler adult intelligence test scale is composed of a number of subtests. on one subtest, the raw scores have a mean of 35 and a standard deviation of 6. assuming these raw scores form a normal distribution:
a) what proportion of raw scores are between 28 and 38?
b) what proportion of raw scores are between 41 and 44?
c) what number represents the 65th percentile (what number separates the lower 65% of the distribution)?
d) what number represents the 90th percentile?
Part (a)
Step1: Calculate z - scores for 28 and 38
The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 35$ (mean) and $\sigma=6$ (standard deviation).
For $x = 28$:
$z_1=\frac{28 - 35}{6}=\frac{- 7}{6}\approx - 1.17$
For $x = 38$:
$z_2=\frac{38 - 35}{6}=\frac{3}{6}=0.5$
Step2: Find the area between the z - scores
We use the standard normal distribution table (or z - table). The area to the left of $z=-1.17$ is $A_1 = 0.1210$ (from z - table). The area to the left of $z = 0.5$ is $A_2=0.6915$ (from z - table).
The proportion of scores between 28 and 38 is $A_2 - A_1=0.6915 - 0.1210 = 0.5705$
Part (b)
Step1: Calculate z - scores for 41 and 44
For $x = 41$:
$z_1=\frac{41 - 35}{6}=\frac{6}{6}=1$
For $x = 44$:
$z_2=\frac{44 - 35}{6}=\frac{9}{6}=1.5$
Step2: Find the area between the z - scores
The area to the left of $z = 1$ is $A_1 = 0.8413$ (from z - table). The area to the left of $z = 1.5$ is $A_2=0.9332$ (from z - table).
The proportion of scores between 41 and 44 is $A_2 - A_1=0.9332 - 0.8413=0.0919$
Part (c)
Step1: Find the z - score corresponding to the 65th percentile
We want to find the z - score $z$ such that $P(Z < z)=0.65$. From the z - table, the z - score corresponding to an area of 0.65 (or the closest value) is $z\approx0.39$ (since the area to the left of $z = 0.39$ is approximately 0.6517).
Step2: Calculate the x - value (raw score)
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$:
$x=\mu+z\sigma$
Substitute $\mu = 35$, $z = 0.39$ and $\sigma = 6$:
$x=35+0.39\times6=35 + 2.34=37.34$
Part (d)
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s:
a) The proportion is $\boldsymbol{0.5705}$
b) The proportion is $\boldsymbol{0.0919}$
c) The 65th percentile is $\boldsymbol{37.34}$ (or approximately 37.3)
d) The 90th percentile is $\boldsymbol{42.68}$ (or approximately 42.7)