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Question

Find the area under the standard normal curve between z = 1.6 and z = 2.6.

0.9452

0.9953

0.6800

0.0501

A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61,400 with a standard deviation of $2,200. Find a 95% confidence interval for the true mean annual income of the business’ customers.

IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual's IQ score is found to be 120. Find the z-score corresponding to this value.

-0.67

0.67

1.33

-1.33

Answer 1

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a) P(1.6<Z<2.6)=P[Z<2.6]-P[Z<1.6]

=0.9953-0.9452

=0.0501

b) n=220

xbar=61400

s=2200

alpha=0.05

First of all, we know the distribution is normal, since the sample size is large (>30). That means our test statistic is either Z or t. Since the standard deviation was estimated from a sample, the population standard deviation is unknown. This means we will use the t-statistic.

So, the confidence interval will be of the form:

mean +/- t(1-alpha/2;n-1) x s/sqrt(n)

ie. 61400 +/- t(0.975;219) x 2200/sqrt(220)

ie. 61400 +/- 0.8347 x 148.324

ie. 61400 +/- 124

ie. (61276,61524)

c) Z=(x-mu)/sigma

=(120-100)/15

=20/15

=1.33

Answer 2

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