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STatiana [176]
3 years ago
9

Joanna set a goal to drink more water daily. The number of ounces of water she drank each of the last seven days is shown below.

Mathematics
1 answer:
DaniilM [7]3 years ago
6 0

(a) Data with the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57,82],  

Sorted data:  [50,57,58,60,64,64,68,82]

Sample size = 8 (even)

mean            = 62.875

median         = (60+64)/2 = 62

1st quartile   = (57+58)/2 = 57.5

3rd quartile  = (64+68)/2 =  66

IQR = 66 - 57.5 = 8.5

(b) Data without the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57]

Sorted data:  [50,57,58,60,64,64,68]

Sample size = 7 (odd)

mean            = 60.143

median         = 60

1st quartile   = 57

3rd quartile = 64

IQR = 64 -57 = 7

Answers:

1. The average is the same with or without the 8th day's data.  FALSE

2. The median is the same with or without the 8th day's data.  FALSE

3. The IQR decreases when the 8th day is included.                  FALSE

4. The IQR increases when the 8th day is included.                   TRUE

5. The median is higher when the 8th day is included.              TRUE

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"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and f
yuradex [85]

Answer:

a

   P(X = 4 ) = 0.1876

b

   P(X \le 4)  =  0.8358

Step-by-step explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

X \  \ ~ Bin (n , p )

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

P(X = 4 ) =  ^nC_4 * p^4 * (1 - p)^{n-4}

=> P(X = 4 ) =  ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}

Here C stand for combination

=> P(X = 4 ) = 0.1876

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as

P(X \le 4) =  [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]

=>   P(X \le 4)  =  0.8358

4 0
3 years ago
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