A technician plans to test a certain type of resin developed in a laboratory to determine the nature of the time it takes before bonding takes place. It is known that the time to bonding is normally distributed with a mean of 4.5 hours and a standard deviation of 1.5 hours. It will be considered an undesirable product if the bonding time is either less than 2 hours or more than 6 hours. a. What is the probability that the bonding time will be less than 2 hours? b. What is the probability that the bonding time will be more than 6 hours? c. How often would the performance be considered undesirable (in a value of probability)? d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?

Step-by-step explanation:

Hello!

The objective is to test if the new type of resin has a bonding time between 2 and 6 hs, less than 2 hs or more than 6 hs makes it undesirable.

The bonding time of resin is normally distributed with mean μ= 4.5 hs and standard deviation δ= 1.5 hs.

a. What is the probability that the bonding time will be less than 2 hours?

Symbolically:

P(X<2)

Using thestandard normal distribution you have to standardize this value to reach the corresponding probability using Z= (X-μ)/δ ~N(0;1)

P(Z<(2-4.5)/1.5)= P(Z<-1.67)= 0.047

b. What is the probability that the bonding time will be more than 6 hours?

P(X>6)= 1 - P(X≤6)

1 - P(Z≤(6-4.5)/1.5)= 1 - P(Z≤1)= 1 - 0.841= 0.159

c. How often would the performance be considered undesirable (in a value of probability)?

The performance will be considered undesirable when the bonding time is less than 2 hs or when the time is higher than 6 hs,

P(X<2) + P(X>6)= 0.047 + 0.159= 0.206

20.6% of the new resin will be considered undesirable.

d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?

I'll start working with the standard normal distribution and then transform the values to X.

Considering that the mean if the standard normal distribution is zero and that the distribution is symmertical regarding it's mean.

You need to look for two values of Z equidistant from the mean that surround 1-α: 0.95 of the distribution, leaving the remaining α: 0.05 outside, equally distributed in two tails α/2: 0.025 (See attachment)

These two value would be the same but with different sign, symbolically:

P(-d≤Z≤d)= 0.95

We can  say that the accumulated probability until "d" is

P(Z≤d)= (1-α)+(α/2)

P(Z≤d)= 0.95+0.025= 0.975

And the accumulated  until "-d" is:

P(Z≤-d)= α/2

P(Z≤-d)= 0.025

Now you look in the table of the standard normal distribution for the corresponding Z values:

P(Z≤d)= 0.975 ⇒ d= 1.965

P(Z≤-d)= 0.025 ⇒ -d= -1.965

Now for each value you have to use the variable information to reverse the standardization and reach the corresponding bonds of the interval:

Upper bond:

Z= (d-μ)/δ

d= (Z*δ)+μ

d= (1.965*1.5)+4.5

d= 7.45

Lower bond

Z= (-d-μ)/δ

-d= (Z*δ)+μ

-d= (-1.965*1.5)+4.5

-d= 1.55

95% of the bonding time will be between 1.55 and 7.45 hs

I hope it helps!

180

step-by-step explanation: