The excel command normdist (x,mean,stdDev,1) returns the value of P(X lessthanor

Question

The excel command normdist (x,mean,stdDev,1) returns the value of P(X lessthanorequal to x) where X is normally distributed with the given mean and standard deviation. Use the NormDist Function to answer the following:

The mean weight of male students at a certain university is 164 Pounds with a standard deviation of 9.4 Pounds.

1. What is the probability that a male student weighs at MOST 178 pounds?

2. What is the probability that a male student weighs at least 158 Pounds?

3. What is the probability that a male student weighs between 171 and 191 pounds?

Please Write out excel formula and equation. I am unsure how to type this into excel.. Thanks. Need HELP FAST

Explanation / Answer

THE MEAN WEIGHT = 164

STANDARD DEVIATION = 9.4

NOW WE WILL USE THE NORMAL DISTRIBUTION

THE FORMULA TO BE USE

Z = (X-MEAN)/STANDARD DEVIATION

A) P(X<=178) =

For x = 178, the z-value z = (178 - 164) / 9.4 = 1.48

Hence P(x < 178) = P(z < 1.48), now from the z table we will take the value of z score = 1.48

And that value will be the probability required.

= [area to the left of 1.48] = 0.9306

B) P(X>=158) =

For x = 158, z = (158 - 164) /9.4 = - 0.63

Hence P(x > 158) = P(z > -0.63) = [total area] - [area to the left of -0.63]

1 - [area to the left of -0.63]

now from the z table we will take the value of z score = -0.63

    = 1 - 0.2643 = 0.7357

C) P(171<X<191) =

For x = 171 , z = (171 - 164) / 9.4 = 0.74 and for x = 191, z = (191 - 164) / 9.4 = 2.87

Hence P(171 < x < 191) = P(2.87 > z > 0.74) = [area to the left of z = 2.87] - [area to the left of 0.74]

= 0.9989 - 0.7995 = 0.1994

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