Question

Sizes of apartments in a large city follow a normal distribution with mean 946 square feet and standard deviation 160 square feet. what is the 60th percentile of apartment sizes in the city?

(a) 986 square feet
(b) 966 square feet
(c) 996 square feet
(d) 906 square feet
(e) 976 square feet

Answer 1

The 60th percentile of apartment sizes in the city is approximately 996 square feet. The option (b) is correct.

Step-by-step explanation:

To find the 60th percentile of the apartment sizes, we can use the Z-score formula and standard normal distribution tables. The Z-score is given by the formula:

`\[ Z = (X - \mu)/(\sigma) \]`

Where
`\(X\)` is the variable of interest (apartment size in this case), \(\mu\) is the mean, and
`\(\sigma\)` is the standard deviation. The 60th percentile corresponds to a Z-score of approximately 0.2533 from the standard normal distribution table.

Using the given information
`(\(\mu = 946\)` and
`\(\sigma = 160\))`, we can rearrange the formula to solve for
`\(X\)`:

`\[ X = Z \cdot \sigma + \mu \]`

Substituting the values:

`\[ X = 0.2533 \cdot 160 + 946 \]`

`\[ X \approx 996 \]`

Therefore, the 60th percentile of apartment sizes in the city is approximately 996 square feet.

This means that 60% of the apartment sizes in the city are below 996 square feet, based on the given mean and standard deviation. Therefore the correct answer is (c) 996 square feet.