Question

In a family with 6 children, the number of children with blue eyes is a binomial random variable with n = 6 and p = 0.23. What is the expected number of children with blue eyes?

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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 60 ounces and a standard deviation of 8 ounces.

Use the Empirical Rule.

Suggestion: sketch the distribution in order to answer these questions.

a) 95% of the widget weights lie between  and  ounces.

b) What percentage of the widget weights lie between 52 and 76 ounces?  %

c) What percentage of the widget weights lie below 84 ounces?  %

Answer 1

Solution:

1. Given X~Binomial(n= 6, p=0.23)
the expected number of children with blue eyes, E(X) = n*p = 6*0.23 = 1.38

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2. mean = 60 , s = 8
a) 95% of data lie between 1 standard deviation of mean
= mean +/- 1.96 * std.deviation
= 60 +/- 1.96 *8
= (44.32, 75.68)
95% of the widget weights lie between 44.32 and 75.68

b)
P(52 < x< 76)
= p((52 - 60)/8 <z < (76 - 60)/8)
= P(-1 < z < 2)

P(52 < x< 76) = P(-1 < z < 2) = 0.8185 by using standard normal tablr

c) P(x > 36)

z = (x -mean)/s
= (84 - 60)/8
= 3

P(x >84) = P(z > 3) = 0.9987 by using standard normal table