Answer to Question #349152 in Statistics and Probability for nurul

Question

Americans ate an average of 25.7 pounds of confectionery products each last year and

spent an average of $61.50 per person doing so. If the standard deviation of consumption is 3.75 pounds and the standard deviation of the amount spent is %5.89, find the following:

a. The probability that the sample mean confectionary consumption for a random sample

of 40 American consumers was greater than 27 pounds.

b. The probability that for a random sample of 50, the sample mean for confectionary

spending exceeded $60.00

Answer 1

Solution:

Let's denote given values:

$\mu1=25.7pounds;$

$\mu2=61.5$ $;

$\sigma1=3.75pounds;$

$X1=27pounds$ -sample value for a);

$X2=60$ $ - sample values for b);

$\sigma2=5.89$ $;

$N1=40$ - sample size for a);

$N2=50$ - sample size for b);

We need to find z-scores:

$z=\frac{X-\mu}{\sigma_e};$ here $\sigma_e=\frac{\sigma}{\sqrt{N}}$ - standard deviation error when N sample numbers;

So,

a) $z1=\frac{X1-\mu1}{\sigma_e1};$ $\sigma_{e1}=\frac{\sigma1}{\sqrt{N1}}=\frac{3.75}{\sqrt{40}}=0.59293;$

$z1=\frac{27-25.7}{0.59293}=2.19;$

Now, we need to check z-table, which $P=0.9857$ for $z1=2.19.$

If we want to find greater then 27 pounds, so

$P1=1-P=1-0.9857=0.0143;$ or $P1=1.43$ %.

b) $z2=\frac{X2-\mu2}{\sigma_e2};$ $\sigma_{e2}=\frac{\sigma2}{\sqrt{N2}}=\frac{5.89}{\sqrt{50}}=0.833;$

$z2=\frac{60-61.5}{0.833}=-1.8;$

$P=0.0359$ for $z2=-1.8.$

If we want to find greater than 60$, so

$P2=1-P=1-0.0358=0.9641;$ or $P2=96.41$ %.

Answer:

a) $P1=1.43$ %.

b) $P2=96.41$ %.