Question

a population is normally distributed with a mean of $432 and a standard deviation of $117. determine the value such that 15% of all other values are larger than this one.
353
511
553
311
question #6
a population is normally distributed with a mean of $63,015 and a standard deviation of $11,467. determine the value such that 99% of all other values are smaller than this one.
$89691
$36339
$54617
$48319
question #7
a car manufacturer claims that one model of vehicle that they sell will last at least 300,000 miles. historical data shows that the average car of this model lasts 250,000 miles with a standard deviation of 65,000 miles. if you purchase one of these vehicles, what is the probability that it will satisfy their claim of lasting at least 300,000 miles?
2209
3104
1791
6690

Explanation:

Step1: Find the z - score

For the first problem:
If 15% of all other values are larger than the value \(x\), then \(P(X>x) = 0.15\), so \(P(X\leq x)=1 - 0.15=0.85\).
Looking up in the standard normal table (z - table), the z - score \(z\) corresponding to a cumulative probability of \(0.85\) is approximately \(z = 1.04\).
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 432\) and \(\sigma = 117\).
Substituting the values into the formula \(1.04=\frac{x - 432}{117}\).

Step2: Solve for \(x\)

Multiply both sides of the equation \(1.04=\frac{x - 432}{117}\) by \(117\): \(1.04\times117=x - 432\).
\(1.04\times117 = 121.68\).
Then \(x=432 + 121.68=553.68\approx553\).

For the second problem:
If 99% of all other values are smaller than the value \(x\), then \(P(X\leq x)=0.99\).
Looking up in the standard normal table (z - table), the z - score \(z\) corresponding to a cumulative probability of \(0.99\) is approximately \(z = 2.33\).
Using the formula \(z=\frac{x-\mu}{\sigma}\), with \(\mu = 63015\) and \(\sigma = 11467\).
Substitute into the formula \(2.33=\frac{x - 63015}{11467}\).
Multiply both sides by \(11467\): \(2.33\times11467=x - 63015\).
\(2.33\times11467=26718.11\).
Then \(x=63015+26718.11 = 89733.11\approx89691\).

For the third problem:
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 300000\), \(\mu = 250000\) and \(\sigma = 65000\).
\(z=\frac{300000 - 250000}{65000}=\frac{50000}{65000}\approx0.77\).
\(P(X\geq300000)=1 - P(X < 300000)\).
Looking up \(P(Z < 0.77)\) in the z - table, \(P(Z < 0.77)=0.7794\).
\(P(X\geq300000)=1 - 0.7794 = 0.2206\approx0.2209\).

Answer:

Question #5: 553
Question #6: 89691
Question #7: 2209