Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation of 0.07 g. If a vending machine will only accept coins weighing between 5.48 g and 5.82 g, what is the probability that a legal quarter will be accepted

0.98046

Step-by-step explanation:

Given:

Here we are required to find

P(5.48 <X<5.82) and between 5.48 and 5.82 we each z-score given by

z = (x - μ) / σ

So 5.48 we have z = -2.714  and 5.82 we have  z = 2.143

Therefore we have the area of interest on the normal distribution chart given by :

P( -2,714 < z < 2.143)

= 1 - P(Z<-2.714) + P(Z >  2.143)

= 1 - P(Z<-2.714) + (1 - P(Z <  2.143))

= 1 - 0.00336 + 1 - 0.98382

= 1 - 0.01954 = 0.98046

98.04%

Step-by-step explanation:

We have to first find the probability that it will be rejected, that is:

P (X <5.48 or X> 5.82)

z = x - m / sd

With a mean of 5.67 and standard deviation of 0.07.

so we have to:

z1 = (5.48 - 5.67) /0.07 = -2.71

In this case we look in the normal distribution table and we have that z is 0.0034

z2 = (5.82 - 5.67) /0.07 = 2.14

In this case we look in the normal distribution table and we have that z is 0.9838

Thus:

P (z <-2.71 or z> 2.14) = P (z <-2.71) + P (z> 2.14)

= P (z <-2.71) + 1 - P (z <2.14)

Superseded:

= 0.0034 + 1 - 0.9838

= 0.0196

And it would be the probability of being rejected, to be accepted it would be:

1 - 0.0196 = 0.9804

In other words, the probability of being accepted is 98.04%

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