Question

A company makes toys whose lives are normally distribution with the mean life, u= 150 hours and standard deviation, a 16 hours.

What % of such toys last between 130 hours and 160 hours?

Answer 1

To find the percentage of toys that last between 130 hours and 160 hours, calculate the z-scores for these values and use the z-table to find the corresponding probabilities. Approximately 63.01% of toys last between 130 and 160 hours.

Step-by-step explanation:

To find the percentage of toys that last between 130 hours and 160 hours, we need to calculate the z-scores for these values and use the z-table to find the corresponding probabilities.

First, calculate the z-score for 130 hours using the formula: z = (130 - mean) / standard deviation.

z = (130 - 150) / 16 = -1.25.

Next, calculate the z-score for 160 hours: z = (160 - 150) / 16 = 0.625.

Using the z-table, we can find the probabilities corresponding to these z-scores.

The probability of a toy lasting less than 130 hours is approximately 0.1056, and the probability of a toy lasting less than 160 hours is approximately 0.7357.

To find the percentage of toys that last between 130 and 160 hours, subtract the probability of a toy lasting less than 130 hours from the probability of a toy lasting less than 160 hours: 0.7357 - 0.1056 = 0.6301.

So, approximately 63.01% of toys last between 130 and 160 hours.