Question

A company manufactures televisions. The average weight of the televisions is 5 pounds with a standard deviation of 0.1 pound. Assuming that the weights are normally distributed, what is the weight that represents the 75th percentile?​

Answer 1

5.0674

Step-by-step explanation:

Given that :

Mean, μ = 5 pounds

Standard deviation, σ = 0.1

Given that weight are normally distributed ;

From the Z table, the Zscore or value for 75th percentile weight is :

P(Z < z) = 0.75

z = 0.674

Using the relation :

Zscore = (x - μ) / σ

x = weight

0.674 = (x - 5) / 0.1

0.674 * 0.1 = x - 5

0.0674 = x - 5

0.0674 + 5 = x - 5 + 5

5.0674 = x

The weight which corresponds to the 75th percentile is 5.0674