1. Let Z be the standard normal variable. Find the value of P(Z > -0.93) a. 0.17

Question

1. Let Z be the standard normal variable. Find the value of P(Z > -0.93)

a. 0.1762
b. 0.3238
c. 0.6619
d. 0.8238

2. Let Z be the standard normal variable. Find the value of P(Z < -1.45)

a. 0.0735
b. 0.0749
c. 0.9251
d. 0.9265

Use the following to answer the following questions.

Use the appropriate normal distribution to approximate the resulting binomial distributions. A
convenience store owner claims that 55% of the people buying from her store, on a certain day of the
week, buy coffee during their visit. A random sample of 35 customers is made. If the store owner's claim
is correct, what is the probability that fewer than 24 customers in the sample buy coffee during their visit
on that certain day of the week?

3. What is the mean(mu) used for the problem?
a. 8.6625
b. 15.7500
c. 19.2500
d. 13.2000
e. 9.5412

4. What is the standard deviation?
a. 2.9432
b. 3.9686
c. 4.3875
d. 3.6332
e. 2.8625


. Find the answer to the problem.
a. 0.9332
b. 0.9251
c. 0.9463
d. 0.8980
e. 0.9525

Explanation / Answer

1) P(Z > -0.93) = 1 - P (Z = -0.93) = 1 - 0.1762 = 0.8238

2)  P(Z < -1.45) = P (Z = -1.45) = 0.0735

3) Proportion of people who visit on a certain day of a week also buy coffee = 55% = 0.55

sample size n = 35

Mean (mu) = Proportion * Sample size = 0.55 * 35 = 19.25

Ans) c 19.2500

4) Std dev = sqrt [P * (1 - P) * n ] = sqrt[0.55 * 0.45 * 35] = 2.9432

Ans) a. 2.9432

5) The probability is given by nCx * Px * ( 1- P )n - x

n =35, x = no of successes = 24, P = 0.55

Z = (X bar - mu)/stdev = (24 - 19.25)/2.94 = 1.61

Probability for (X<24)   = P(Z<1.61) = 0.9463

Ans) c. 0.9463

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