Question

1. Based from record, the mean weight of the newborn babies is 4.8 lbs with a standard deviation of 0.75 lbs. Out of 60 new born babies, find:

a. The probability that the sample mean is less than 4.6 lbs

b.The probability that the sample mean is equal to 5.0 lbs

c. The probability that the sample mean is greater than 5.1 lbs

d. The probability that the sample mean is between 4.5 to 4.9 lbs

2. Scanned imagine occupied an average of 0.6 megabytes of memory with a standard deviation of 0.4 megabytes. If you plan to upload 80 images on your website,what is the probability that their total size is between 47 megabytes and 56 megabytes

Answer 1

1. Let $X=$ the sample mean: $X\sim N(\mu, \sigma^2/n)$

Given $\mu=4.8, \sigma=0.75, n=60.$

a.

$$P(X<4.6)=P(Z<\dfrac{4.6-4.8}{0.75/\sqrt{60}})$$

$$\approx P(Z<-2.0656)\approx0.0194$$

b.

$$P(X=5.0)=0$$

c.

$$P(X>5.1)=1-P(X\le5.1)$$

$$=1-P(Z\le\dfrac{5.1-4.8}{0.75/\sqrt{60}})$$

$$\approx1-P(\le3.0984)\approx0.0010$$

d.

$$P(4.5<X<4.9)=P(X<4.9)-P(X\le4.5)$$

$$=P(Z<\dfrac{4.9-4.8}{0.75/\sqrt{60}})-P(Z\le\dfrac{4.5-4.8}{0.75/\sqrt{60}})$$

$$\approx P(Z<1.0328)-P(\le-3.0984)$$

$$\approx0.8492-0.0010\approx0.8482$$

2. Let $X=$ the total size: $X\sim N(n\mu, n\sigma^2)$

Given $\mu=0.6, \sigma=0.4, n=80.$

$$P(47<X<56)=P(X<56)-P(X\le47)$$

$$=P(Z<\dfrac{56-48}{0.4\sqrt{80}})-P(Z\le\dfrac{47-48}{0.4/\sqrt{80}})$$

$$\approx P(Z<2.2361)-P(\le-0.2795)$$

$$\approx0.9873-0.3899\approx0.5974$$