Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50, with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. (a)What is the likelihood the sample mean is at least $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability (b)What is the likelihood the sample mean is greater than $22.50 but less an $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability (c)Within what limits will 90% of the sample means occur? (Round your answers to 2 decimal places.) The limits are and

Answer 1

Standard normal distribution of a random variable x,

1 -

where z is the Standard normal distribution value, $\mu$ = Mean of the sample = 23.5, $\sigma$ = Standard deviation = 5

a) x has to be atleast 25, means X $\geq$ 25

z = (25-23.5) / 5 = 0.3

Using the NORMSDIST function in excel, =NORSDIST(0.3) = 0.617911

PIZZ

This 0.617911 is the probability that x<25, as shown by the blue shaded region.

Hence to get probability that mean is atleast 25 = 1 - 0.617911 = 0.382089 = 38.2089%

b) Probability that 22.5<x<25

First let us calculate probability of x>22.5

z = (22.5-23.5) / 5 = -0.2

Using the NORMSDIST function in excel, =NORSDIST(-0.2) = 0.42074

(25Z)d

This 0.42074 is the probability that x<22.5, as shown by the blue shaded region.

Hence to get probability that x > 22.5 = 1 - 0.42074 = 0.579260.......(1)

We already know the probability that x > 25 = 0.382089........(2)

So to get Probability that 22.5<x<25, we deduct (2) from (1) & get 0.197171 = 19.7171%

c)

Confidenceinterval = Mean Ez*

z_0.05 = 1.645 [Since given, 90% confidence, area under each tail = 5%. So we have to look for 95% confidence in the left tail in the standard distribution table]

x| 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 10.0 0.5000 0.5040 | 0.5080|0.5120|0.5160 | 0.5199|0.5239 10. 1 0.5398 0.5438|0.5478|0.5517|0.5557 0.5596|0.5636 10.2 0.5793 0.5832 | 0.5871| 0.5910| 0.5948|0.5987 0.6026 0.3 0.6179 | 0.6217 0.6255 0.6293 0.6331| 0.6368 0.6406 10.4 0.6554 0.6591| 0.6628 0.6664 | 0.6700| 0.6736 0.6772 | 0.5 0.6915 0.6950 | 0.6985 | 0.7019 0.7054| 0.7088 0.7123 10.6 0.7257 0.729110.73240.7357 10.7389 | 0.7422 | 0.7454 | 0.7|0.7580 0.7611| 0.7642 | 0.7673 | 0.7704 0.7734 0.7764 | 0.8 0.7881| 0.7910| 0.7939 | 0.7967 | 0.7995 | 0.8023 0.8051 10.9 0.8159 0.8186 0.8212 0.8238|0.8264 0.8289 0.8315 | 1.0 0.8413| 0.8438 | 0.8461| 0.8485 | 0.8508| 0.8531 0.8554 | 1. 10.8643 | 0.8665 | 0.8686| 0.8708 | 0.8729 | 0.8749 0.8770 | 1.2 | 0.8849 | 0.8869 | 0.8888 | 0.8907] 0.8925 | 0.8944 0.8962 1.3 0.9032 | 0.9049 | 0.9066 | 0.9082 | 0.9099 | 0.9115 0.9131 1.4 0.9192 0.920709222|09236|0.9251| 0.9265 0.9279 11.5 | 0.9332 | 0.9345 | 0.9357 0.9370| 0.9382| 0.9394 0.9406 | 1.6 0.9452 0.9463 | 0.9474| 0.9484 0.9495 | 0.9505 0.9515

Mean = 23.5, Std dev = 5, n = 50

Substituting above values, we get 23.5$\pm$ 1.163191

So confidence interval: 23.5-1.163191, 23+1.163191

= 22.3368, 24.6632