Question

The weight of an almond varies with mean 0.048 ounce and standard deviation 0.011 ounce.

a. What is the probability that the total weight (of a random sample) of 60 almonds is less than 3 ounces? Give your answer to 2 decimal places.

b. Determine the minimum sample size (the number of almonds) needed to have the probability of at least 0.80 that the total weight is greater than 16 ounces. Enter an integer below. (Note: You must round up, e.g. 231.23671 → 232.)

Please show work

Answer 1

mean =0.048 ounce

standard deviation = 0.011 ounce

mean weight of 60 sample= n*mean=60*0.048=2.88

var (total) =n*std dev²= 60*0.011^2=0.00726

std dev(total) = 0.085

a)

Z = (x-mean) / SE =(3-2.88)/0.085 = 1.4118

P(X<3)=P(Z<1.4118) = 0.92099

b)

z -score corresponding to probbaility 0.80 is

P(Z>-0.84)=0.80 { =normsinv(0.80) for z score}

P(X>16) = 0.80

z=(16-n*mean)/std dev* sqrt n = -0.84

(16-n*0.048) / 0.011*sqrt n = -0.84

16-0.048n = -0.00924 *sqrt n

0.048n-0.00924 * sqrt n -16 =0

n=[ 1.53609 + (0.00026)^0.5 ] /0.004608 = 336.85

so, n=337