The time required to assemble an electronic component is normally distributed wi

Question

The time required to assemble an electronic component is normally distributed with a mean and a standard deviation of 16 minutes and 4 minutes, respectively. Use Table 1. a. Find the probability that a randomly picked assembly takes between 10 and 20 minutes. (Round "z" value to 2 decimal places and final answer to 4 decimal places.) Probability b. It is unusual for the assembly time to be above 24 minutes or below 6 minutes. What proportion of assembly times fall in these unusual categories? (Round "z" value to 2 decimal places and final answer to 4 decimal places.) Proportion of assembly times

Explanation / Answer

Solution:-

Mean = 16 minutes

Standard deviation = 4 minutes

a). P(10 < X < 20)

= 16 and = 4, we have:

P ( 10 < X < 20 ) = P ( 1016 < X < 2016 )

=P ( 1016 / 4 < X / < 2016 / 4)

Since Z = x / , 1016 / 4 = 1.5 and 2016 / 4 = 1, we have:

P ( 10 < X < 20 ) = P ( 1.5 < Z < 1 )

Use the standard normal table to conclude that:

P ( 1.5 < Z < 1 ) = 0.7745

b). Above 24 minutes

P ( X > 24 ) = P ( X > 2416 ) = P ( X / > 2416 / 4)

Since Z = x / and 2416 / 4 = 2, we have:

P ( X > 24 ) = P ( Z > 2 )

Use the standard normal table to conclude that:

P (Z > 2) = 0.0228

Below 6 minutes:-

Since = 16 and = 4, we have:

P ( X < 6 ) = P ( X < 616 ) = P (X / < 616 / 4)

Since x / =Z and 616 / 4 = 2.5, we have:

P (X < 6) = P (Z < 2.5)

Use the standard normal table to conclude that:

P (Z < 2.5) = 0.0062

P(24 < Z < 6) = 0.0228 + 0.0062 = 0.029 = 0.03

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