If z is a standard normal variable, find the probability 4) The probability that

Question

If z is a standard normal variable, find the probability 4) The probability that z is greater than -1.82 Find the indicated value. 5) 20.02 Provide an appropriate response. 6) 1Q tests are designed so that the mean I10 score is 100, with a standard deviation of 15. This means that a measure of 1Qs across the overall population looks like a bell curve or normal distribution. Find P35, which is the IQ score separating the bottom 35% from the top 68%. 7) 1Q tests are designed so that the mean 1Q score is 100, with a standard deviation of 15. This means that a measure of IQs across the overall population looks like that a randomly selected adult has an 1Q between 90 and 120 (somewhere in the range of normal to bright normal). a bell curve or normal distribution. Find the probability Find the indicated probability 5) The People Mover will only accept tokens weighing between 5.48 g and 582 g. Assume that the weights of the tokens are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. What percentage of real tokens will be rejected? Is this a problem? Explain. B-2

Explanation / Answer

4.
P(z > -1.82)
= 1 - P(z < -1.82)
= 1 - 0.0344 .. using left tailed z table
= 0.9656

5.
z(0.02) = -2.0537

6.
mean = 100, sd = 15
z-value = -0.3853 .. (using z-table)
xbar = 100 - 0.3853*15 = 94.2205

P(35) = 94.2205

7.
P(90 < X < 120)
= P((90 - 100)/15 < z < (120 - 100)/15))
= P(-0.6667 < z < 1.3333)
= 0.9088 - 0.2525
= 0.6563

8.
P(X < 5.48) + P(X > 5.82)
= P(z < (5.48 - 5.67)/0.07) + P(z > (5.82 - 5.67)/0.07)
= P(z < -2.7143) + P(z > 2.1429)
= 0.0033 + 0.0161
= 0.0194

Hence 1.94% of real tokens will be rejected.

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