Question

The amount of time a shopper spends in the supermarket is normally distributed with a mean of 45 minutes with a standard deviation of 12 minutes. If 10 shoppers are selected, how many of them will spend less than 35 minutes? Round your answer to 3 decimal places. Do not include units in your answer. 4 points

QUESTION 13 The amount of time a shopper spends in the supermarket is normally distributed with a mean of 45 minutes with a standard deviation of 12 minutes. Your spouse believes you are in the top 10% of times shoppers spend in supermarkets. How many minutes (at least) does your spouse believe you spend in the supermarket? Round your answer to 2 decimal places. Do not include units in your answer. 5 points

QUESTION 14 The amount of time a shopper spends in the supermarket is normally distributed with a mean of 45 minutes with a standard deviation of 12 minutes. Ten shoppers are randomly selected.What is the mean of the sample means? Round your answer to 2 decimal places. Do not include units in your answer.

Answer 1

Solution :

Given that ,

mean = = 45

standard deviation = = 12

P(x < 35 ) = P[(x - ) / < ( 35 - 45 ) / 12]

= P(z < -0.83 )

Using z table,

= 0.2033 * 10

= 2.033

Answer = 2.033

Q - 13

mean = = 45

standard deviation = = 12

The z - distribution of the 10% is ,

P(Z z) = 10%

= 1 - P(Z z ) = 0.10

= P(Z ) = 1 - 0.10

= P(Z z ) = 0.90

= P(Z 1.282 ) = 0.90

z = 1.282

Using z-score formula,

x = z * +

x = 1.282 * 12 + 45

x = 60.384

Answer = 60.38

Q - 14

mean = = 45

standard deviation = = 12

n = 10

= 45

= / n = 12 / 10 = 3.79

The mean of the sample mean = 45

The standard deviation of the sample mean = 3.79